v = x t = x 2 x 1 t 2 t 1 The average speed of the particle is absolute value of the average velocity and is given Distance travelled t

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "v = x t = x 2 x 1 t 2 t 1 The average speed of the particle is absolute value of the average velocity and is given Distance travelled t"

Transcription

1 Chapter 2 Motion in One Dimenion 2.1 The Important Stuff Poition, Time and Diplacement We begin our tudy of motion by conidering object which are very mall in comparion to the ize of their movement through pace. When we can deal with an object in thi way we refer to it a a particle. In thi chapter we deal with the cae where a particle move along a traight line. The particle location i pecified by it coordinate, which will be denoted by x or y. A the particle move, it coordinate change with the time, t. The change in poition from x 1 to x 2 of the particle i the diplacement x, with x = x 2 x Average Velocity and Average Speed When a particle ha a diplacement x in a change of time t, it average velocity for that time interval i by v = x t = x 2 x 1 t 2 t 1 (2.1) The average peed of the particle i abolute value of the average velocity and i given = Ditance travelled t (2.2) In general, the value of the average velocity for a moving particle depend on the initial and final time for which we have found the diplacement Intantaneou Velocity and Speed We can anwer the quetion how fat i a particle moving at a particular time t? by finding the intantaneou velocity. Thi i the limiting cae of the average velocity when the time 27

2 28 CHAPTER 2. MOTION IN ONE DIMENSION interval t include the time t and i a mall a we can imagine: x v = lim t 0 t = dx dt (2.3) The intantaneou peed i the abolute value (magnitude) of the intantaneou velocity. If we make a plot of x v. t for a moving particle the intantaneou velocity i the lope of the tangent to the curve at any point Acceleration When a particle velocity change, then we way that the particle undergoe an acceleration. If a particle velocity change from v 1 to v 2 during the time interval t 1 to t 2 then we define the average acceleration a v = x t = x 2 x 1 t 2 t 1 (2.4) A with velocity it i uually more important to think about the intantaneou acceleration, given by v a = lim t 0 t = dv (2.5) dt If the acceleration a i poitive it mean that the velocity i intantaneouly increaing; if a i negative, then v i intantaneouly decreaing. Oftentime we will encounter the word deceleration in a problem. Thi word i ued when the ene of the acceleration i oppoite that of the intantaneou velocity (the motion). Then the magnitude of acceleration i given, with it direction being undertood Contant Acceleration A very ueful pecial cae of accelerated motion i the one where the acceleration a i contant. For thi cae, one can how that the following are true: v = v 0 + at (2.6) x = x 0 + v 0 t at2 (2.7) v 2 = v a(x x 0 ) (2.8) x = x (v 0 + v)t (2.9) In thee equation, we mean that the particle ha poition x 0 and velocity v 0 at time t = 0; it ha poition x and velocity v at time t. Thee equation are valid only for the cae of contant acceleration.

3 2.2. WORKED EXAMPLES Free Fall An object toed up or down near the urface of the earth ha a contant downward acceleration of magnitude 9.80 m 2. Thi number i alway denoted by g. Be very careful about the ign; in a coordinate ytem where the y axi point traight up, the acceleration of a freely falling object i a y = 9.80 m 2 = g (2.10) Here we are auming that the air ha no effect on the motion of the falling object. For an object which fall for a long ditance thi can be a bad aumption. Remember that an object in free fall ha an acceleration equal to 9.80 m 2 while it i moving up, while it i moving down, while it i at maximum height... alway! 2.2 Worked Example Average Velocity and Average Speed 1. Boton Red Sox pitcher Roger Clemen could routinely throw a fatball at a horizontal peed of 160 km. How long did the ball take to reach home plate 18.4m away? [HRW5 2-4] hr We aume that the ball move in a horizontal traight line with an average peed of 160 km/hr. Of coure, in reality thi i not quite true for a thrown baeball. We are given the average velocity of the ball motion and alo a particular diplacement, namely x = 18.4 m. Equation 2.1 give u: v = x t = t = x v But before uing it, it might be convenient to change the unit of v. We have: v = 160 km hr ( ) ( ) 1000m 1hr = 44.4 m 1km 3600 Then we find: t = x v = 18.4m 44.4 m = The ball take econd to reach home plate. 2. Taking the Earth orbit to be a circle of radiu km, determine the peed of the Earth orbital motion in (a) meter per econd and (b) mile per econd. [Wolf 2-18]

4 30 CHAPTER 2. MOTION IN ONE DIMENSION (a) Thi i not traight line motion of coure, but we can ill find an average peed by dividing the ditance traveled (around a circular path) by the time interval. Here, the ditance traveled by the Earth a it goe once around the Sun i the circumference of the orbit, C = 2πR = 2π( km) = km = m and the time interval over which that take place i one year, o the average peed i ( ) (3600 ) 24hr 1yr = day = day 1hr = C t = m = m (b) To convert thi to mi, ue 1mi = 1.609km. Then Acceleration = ( m ) ( ) 1mi = 18.6 mi m 3. An electron moving along the x axi ha a poition given by x = (16te t )m, where t i in econd. How far i the electron from the origin when it momentarily top? [HRW6 2-20] To find the velocity of the electron a a function of time, take the firt derivative of x(t): v = dx dt = 16e t 16te t = 16e t (1 t) m again where t i in econd, o that the unit for v are m. Now the electron momentarily top when the velocity v i zero. From our expreion for v we ee that thi occur at t = 1. At thi particular time we can find the value of x: x(1) = 16(1)e 1 m = 5.89m The electron wa 5.89m from the origin when the velocity wa zero. 4. (a) If the poition of a particle i given by x = 20t 5t 3, where x i in meter and t i in econd, when if ever i the particle velocity zero? (b) When i it acceleration a zero? (c) When i a negative? Poitive? (d) Graph x(t), v(t), and a(t). [HRW5 2-28]

5 2.2. WORKED EXAMPLES 31 (a) From Eq. 2.3 we find v(t) from x(t): v(t) = dx dt = d dt (20t 5t3 ) = 20 15t 2 where, if t i in econd then v will be in m. The velocity v will be zero when which we can olve for t: 20 15t 2 = 0 15t 2 = 20 = t 2 = = (The unit 2 were inerted ince we know t 2 mut have thee unit.) Thi give: t = ±1.15 (We hould be careful... t may be meaningful for negative value!) (b) From Eq. 2.5 we find a(t) from v(t): a(t) = dv dt = d dt (20 15t2 ) = 30t where we mean that if t i given in econd, a i given in m 2. From thi, we ee that a can be zero only at t = 0. (c) From the reult i part (b) we can alo ee that a i negative whenever t i poitive. a i poitive whenever t i negative (again, auming that t < 0 ha meaning for the motion of thi particle). (d) Plot of x(t), v(t), and a(t) are given in Fig In an arcade video game a pot i programmed to move acro the creen according to x = 9.00t 0.750t 3, where x i ditance in centimeter meaured from the left edge of the creen and t i time in econd. When the pot reache a creen edge, at either x = 0 or x = 15.0cm, t i reet to 0 and the pot tart moving again according to x(t). (a) At what time after tarting i the pot intantaneouly at ret? (b) Where doe thi occur? (c) What i it acceleration when thi occur? (d) In what direction i it moving jut prior to coming to ret? (e) Jut after? (f) When doe it firt reach an edge of the creen after t = 0? [HRW5 2-31] (a) Thi i a quetion about the intantaneou velocity of the pot. To find v(t) we calculate: v(t) = dx dt = d dt (9.00t 0.750t3 ) = t 2 where thi expreion will give the value of v in cm when t i given in econd.

6 32 CHAPTER 2. MOTION IN ONE DIMENSION x, (m) t v, (m/) t a, (m/ 2 ) t Figure 2.1: Plot of x(t), v(t), and a(t) for Example 4.

7 2.2. WORKED EXAMPLES 33 We want to know the value of t for which v i zero, i.e. the pot i intantaneouly at ret. We olve: t 2 = 0 = t 2 = = (Here we have filled in the proper unit for t 2 ince by lazine they were omitted from the firt equation!) The olution to thi equation are t = ±2.00 but ince we are only intereted in time after the clock tart at t = 0, we chooe t = (b) In thi part we are to find the value of x at which the intantaneou velocity i zero. In part (a) we found that thi occurred at t = 3.00 o we calculate the value of x at t = 2.00: x(2.00) = 9.00 (2.00) (2.00) 3 = 12.0cm (where we have filled in the unit for x ince centimeter are implied by the equation). The dot i located at x = 12.0cm at thi time. (And recall that the width of the creen i 15.0cm.) (c) To find the (intantaneou) acceleration at all time, we calculate: a(t) = dv dt = d dt ( t2 ) = 4.50t where we mean that if t i given in econd, a will be given in m 2. At the time in quetion (t = 2.00 ) the acceleration i a(t = 2.00) = 4.50 (2.00) = 9.00 that i, the acceleration at thi time i 9.00 m 2. (d) From part (c) we note that at the time that the velocity wa intantaneouly zero the acceleration wa negative. Thi mean that the velocity wa decreaing at the time. If the velocity wa decreaing yet intantaneouly equal to zero then it had to be going from poitive to negative value at t = So jut before thi time it velocity wa poitive. (e) Likewie, from our anwer to part (d) jut after t = 2.00 the velocity of particle had to be negative. (f) We have een that the dot never get to the right edge of the creen at x = 15.0 cm. It will not revere it velocity again ince t = 2.00 i the only poitive time at which v = 0. So it will keep moving to back to the left, and the coordinate x will equal zero when we have: x = 0 = 9.00t 0.750t 3 Factor out t to olve: t( t 2 ) = 0 = { t = 0 or ( t 2 ) = 0 otherwie.

8 34 CHAPTER 2. MOTION IN ONE DIMENSION x, cm t, Figure 2.2: Plot of x v t for moving pot. Ignore the part where x i negative! The firt olution i the time that the dot tarted moving, o that i not the one we want. The econd cae give: which give ( t 2 ) = 0 = t 2 = = t = 3.46 ince we only want the poitive olution. So the dot return to x = 0 (the left ide of the creen) at t = If we plot the original function x(t) we get the curve given in Fig. 2.2 which how that the pot doe not get to x = 15.0 cm before it turn around. (However a explained in the problem, the curve doe not extend to negative value a the graph indicate.) Contant Acceleration 6. The head of a rattlenake can accelerate 50 m 2 in triking a victim. If a car could do a well, how long would it take to reach a peed of 100 km hr from ret? [HRW5 2-33] Firt, convert the car final peed to SI unit to make it eaier to work with: 100 km ( hr = 100 km ) ( ) ( ) 1000m 1hr = 27.8 m hr 1km 3600 The acceleration of the car i 50 m and it tart from ret which mean that v 2 0 = 0. A we ve found, the final velocity v of the car i 27.8 m. (The problem actually that thi i final

9 2.2. WORKED EXAMPLES 35 peed but if our coordinate ytem point in the ame direction a the car motion, thee are the ame thing.) Equation 2.6 let u olve for the time t: Subtituting, we find v = v 0 + at = t = v v 0 a t = 27.8 m 0 50 m 2 = 0.55 If a car had uch a large acceleration, it would take 0.55 to attain the given peed. 7. A body moving with uniform acceleration ha a velocity of 12.0 cm when it x coordinate i 3.00cm. If it x coordinate 2.00 later i 5.00cm, what i the magnitude of it acceleration? [Ser4 2-25] In thi problem we are given the initial coordinate (x = 3.00cm), the initial velocity (v 0 = 12.0 cm ), the final x coordinate (x = 5.00cm) and the elaped time (2.00). Uing Eq. 2.7 (ince we are told that the acceleration i contant) we can olve for a. We find: Subtitute thing: x = x 0 + v 0 t at2 = 1 2 at2 = x x 0 v 0 t Solve for a: 1 2 at2 = 5.00cm 3.00cm ( 12.0 cm a = 2( 32.0cm) t 2 = 2( 32.0cm) (2.00) 2 ) (2.00) = 32.0cm = 16.0 cm 2 The x acceleration of the object i 16. cm 2. (The magnitude of the acceleration i 16.0 cm 2.) 8. A jet plane land with a velocity of 100 m and can accelerate at a maximum rate of 5.0 m 2 a it come to ret. (a) From the intant it touche the runway, what i the minimum time needed before it top? (b) Can thi plane land at a mall airport where the runway i 0.80 km long? [Ser4 2-31] (a) The data given in the problem i illutrated in Fig The minu ign in the acceleration indicate that the ene of the acceleration i oppoite that of the motion, that i, the plane i decelerating. The plane will top a quickly a poible if the acceleration doe have the value 5.0 m, 2 o we ue thi value in finding the time t in which the velocity change from v 0 = 100 m to v = 0. Eq. 2.6 tell u: t = v v 0 a Subtituting, we find: t = (0 100 m) ( 5.0 m = 20 ) 2

10 36 CHAPTER 2. MOTION IN ONE DIMENSION a = -5.0 m/ m/ v = 0 x Figure 2.3: Plane touche down on runway at 100 m and come to a halt. The plane need 20 to come to a halt. (b) The plane alo travel the hortet ditance in topping if it acceleration i 5.0 m 2. With x 0 = 0, we can find the plane final x coordinate uing Eq. 2.9, uing t = 20 which we got from part (a): x = x (v 0 + v)t = (100 m + 0)(20) = 1000m = 1.0km The plane mut have at leat 1.0km of runway in order to come to a halt afely. 0.80km i not ufficient. 9. A drag racer tart her car from ret and accelerate at 10.0 m for the entire 2 ditance of 400m ( 1 mile). (a) How long did it take the car to travel thi ditance? 4 (b) What i the peed at the end of the run? [Ser4 2-33] (a) The racer move in one dimenion (along the x axi, ay) with contant acceleration a = 10.0 m 2. We can take her initial coordinate to be x 0 = 0; he tart from ret, o that v 0 = 0. Then the location of the car (x) i given by: x = x 0 + v 0 t at2 = = at2 = 1 2 (10.0 m 2 )t 2 We want to know the time at which x = 400m. Subtitute and olve for t: which give 400m = 1 2 (10.0 m 2 )t 2 = t 2 = 2(400m) (10.0 m 2 ) = The car take 8.94 to travel thi ditance. t = (b) We would like to find the velocity at the end of the run, namely at t = 8.94 (the time we found in part (a)). The velocity i: v = v 0 + at = 0 + (10.0 m 2 )t = (10.0 m 2 )t

11 2.2. WORKED EXAMPLES 37 Accelerating region 1.0 cm Path of electron Voltage ource Figure 2.4: Electron i accelerated in a region between two plate, in Example 10. At t = 8.94, the velocity i The peed at the end of the run i 89.4 m. v = (10.0 m 2 )(8.94) = 89.4 m 10. An electron with initial velocity v 0 = m enter a region 1.0cm long where it i electrically accelerated, a hown in Fig It emerge with velocity v = m. What wa it acceleration, aumed contant? (Such a proce occur in the electron gun in a cathode ray tube, ued in televiion receiver and ocillocope.) [HRW5 2-39] We are told that the acceleration of the electron i contant, o that Eq can be ued. Here we know the initial and final velocitie of the electron (v 0 and v). If we let it initial coordinate be x 0 = 0 then the final coordinate i x = 1.0cm = m. We don t know the time t for it travel through the accelerating region and of coure we don t know the (contant) acceleration, which i what we re being aked in thi problem. We ee that we can olve for a if we ue Eq. 2.8: Subtitute and get: v 2 = v a(x x 0 ) = a = v2 v 2 0 2(x x 0 ) a = ( m )2 ( m )2 2( m) = m 2 The acceleration of the electron i m 2 (while it i in the accelerating region).

12 38 CHAPTER 2. MOTION IN ONE DIMENSION 11. A world land peed record wa et by Colonel John P. Stapp when on March 19, 1954 he rode a rocket propelled led that moved down a track at 1020 km. He and the led were brought to a top in 1.4. What acceleration did h he experience? Expre your anwer in g unit. [HRW5 2-41] For the period of deceleration of the rocket led (which lat for 1.4 ) were are given the initial velocity and the final velocity, which i zero ince the led come to ret at the end. Firt, convert hi initial velocity to SI unit: v 0 = 1020 km h = (1020 km h ) ( 10 3 m 1km The Eq. 2.6 give u the acceleration a: Subtitute: ) ( ) 1h = m 3600 v = v 0 + at = a = v v 0 t a = m = m The acceleration i a negative number ince it i oppoite to the ene of the motion; it i a deceleration. The magnitude of the led acceleration i m. 2 To expre thi a a multiple of g, we note that a g = m m 2 = 20.7 o the magnitude of the acceleration wa a = 20.7g. That a lotta g! 12. A ubway train i traveling at 80 km when it approache a lower train 50m h ahead traveling in the ame direction at 25 km. If the fater train begin decelerating at 2.1 m while the lower train continue at contant peed, how oon and h 2 at what relative peed will they collide? [wolf 2-73] m Firt, convert the initial peed of the train to unit of. We find: 80 km h = 22.2 m 25 km h = 6.94 m. The ituation of the train at t = 0 (when the rear train begin to decelerate) i hown in Fig We chooe the origin of the x axi to be at the initial poition of the rear train; then the initial poition of the front train i x = 50m. If we call the x coordinate of the rear train x 1, then ince it ha initial velocity 22.2 m and acceleration 2.1 m 2 (note the minu ign!) the equation for x 1 (t) i x 1 (t) = (22.2 m )t ( 2.1 m 2 )t 2 = (22.2 m )t + ( 1.05 m 2 )t 2 Meanwhile, the front car ha an initial velocity of 6.94 m and no acceleration, o it coordinate (x 2 ) i given by x 2 (t) = 50m + (6.94 m )t

13 2.2. WORKED EXAMPLES 39 a = -2.1 m/ m/ 6.94 m/ 1 2 x 1 50 m x 2 Figure 2.5: Two ubway train in Example 12. The train will collide if there i ever a time at which their coordinate are equal. So we want to ee if there i a t which give the condition: (22.2 m )t + ( 1.05 m 2 )t 2 = 50m + (6.94 m )t Thi i a quadratic equation, for which we can ue the quadratic formula. Neglecting the unit for implicity, we can rearrange the term and rewrite it a 1.05t t + 50 = 0 and the quadratic formula give the anwer a b ± b 2 4ac 2a = ± (15.28) 2 4(1.05)(50) 2(1.05) = { Thi i a little confuing becaue there are two poible anwer! (Both value of t are poitive.) But the anwer we want i the firt one, 4.97 after the colliion, the econd time i not relevant 1. So the train will collide t = 4.97 after the rear car begin to decelerate. At the time we have found, the velocity of the rear train i v = v 0 + at = 22.2 m + ( 2.1 m )(4.97) = 11.8 m 2 and the velocity of the front train remain 6.94 m. So at the time of the colliion, the rear train i going fater by a difference of v = 11.8 m 6.94 m = 4.8 m That i the relative peed at which the colliion take place Free Fall

14 40 CHAPTER 2. MOTION IN ONE DIMENSION y y=50 m v 0 y= 0 m Figure 2.6: Object thrown upward reache height of 50 m. 13. (a) With what peed mut a ball be thrown vertically from ground level to rie to a maximum height of 50m? (b) How long will it be in the air? [HRW5 2-61] (a) Firt, we decide on a coordinate ytem. I will ue the one hown in Fig. 2.6, where the y axi point upward and the origin i at ground level. The ball tart it flight from ground level o it initial poition i y 0 = 0. When the ball i at maximum height it coordinate i y = 50 m, but we alo know it velocity at thi point. At maximum height the intantaneou velocity of the ball i zero. So if our final point i the time of maximum height, then v = 0. So for the trip from ground level to maximum height, we know y 0, y, v and the acceleration a = 9.8 m 2 = g, but we don t know v 0 or the time t to get to maximum height. From our lit of contant acceleration equation, we ee that Equation 2.8 will give u the initial velocity v 0 : Subtitute, and get: v 2 = v a(y y 0 ) = v 2 0 = v 2 2a(y y 0 ) v 2 0 = (0) 2 2( 9.8 m 2 )(50m 0) = 980 m2 2 The next tep i to take the quare root. Since we know that v 0 mut be a poitive number, we know that we hould take the poitive quare root of 980 m2 2. We get: v 0 = +31 m The initial peed of the ball i 31 m (b) We want to find the total time that the ball i in flight. What do we know about the ball when it return to earth and hit the ground? We know that it y coordinate i equal to zero. (So far, we don t know anything about the ball velocity at the the time it return to ground level.) If we conider the time between throwing and impact, then we do know y 0, y, v 0 and of coure a. If we ubtitute into Eq. 2.7 we find: 1 However it would be relevant if the train were on parallel track; then the colliion would not take place and we could find the time at which they were ide-by-ide and their relative velocitie at thoe time.

15 2.2. WORKED EXAMPLES 41 y 8.00 m/ 30.0 m Figure 2.7: Ball i thrown traight down with peed of 8.00 m, in Example = 0 + (31 m )t ( 9.8 m 2 )t 2 It i not hard to olve thi equation for t. We can factor it to give: t[(31 m ) ( 9.8 m 2 )t] = 0 which ha two olution. One of them i imply t = 0. Thi olution i an anwer to the quetion we are aking, namely When doe y = 0? becaue the ball wa at ground level at t = 0. But it i not the olution we want. For the other olution, we mut have: which give (31 m ) ( 9.8 m 2 )t = 0 t = 2(31 m ) 9.8 m 2 = 6.4 The ball pend a total of 6.4 econd in flight. 14. A ball i thrown directly downward with an initial peed of 8.00 m from a height of 30.0m. When doe the ball trike the ground? [Ser4 2-46] We diagram the problem a in Fig We have to chooe a coordinate ytem, and here I will put the let the origin of the y axi be at the place where the ball tart it motion (at the top of the 30m height). With thi choice, the ball tart it motion at y = 0 and trike the ground when y = 30m. We can now ee that the problem i aking u: At what time doe y = 30.0m? We have v 0 = 8.00 m (minu becaue the ball i thrown downward!) and the acceleration of the the ball i a = g = 9.8 m 2, o at any time t the y coordinate i given by y = y 0 + v 0 t at2 = ( 8.00 m )t 1 2 gt2

16 42 CHAPTER 2. MOTION IN ONE DIMENSION y v m Figure 2.8: Student throw her key into the air, in Example 15. But at the time of impact we have y = 30.0m = ( 8.00 m )t 1 2 gt2 = ( 8.00 m )t (4.90 m 2 )t 2, an equation for which we can olve for t. We rewrite it a: (4.90 m )t 2 + (8.00 m )t 30.0m = 0 2 which i jut a quadratic equation in t. From our algebra coure we know how to olve thi; the olution are: t = (8.00 m ) ± (8.00 m )2 4(4.90 m 2 )( 30.0m) 2(4.90 m 2 ) and a little calculator work finally give u: t = { Our anwer i one of thee... which one? Obviouly the ball had to trike the ground at ome poitive value of t, o the anwer i t = The ball trike the ground 1.78 after being thrown. 15. A tudent throw a et of key vertically upward to her orority iter in a window 4.00 m above. The key are caught 1.50 later by the iter outtretched hand. (a) With what initial velocity were the key thrown? (b) What wa the velocity of the key jut before they were caught? [Ser4 2-47] (a) We draw a imple picture of the problem; uch a imple picture i given in Fig Having a picture i important, but we hould be careful not to put too much into the picture; the problem did not ay that the key were caught while they were going up or going down. For all we know at the moment, it could be either one!

17 2.2. WORKED EXAMPLES 43 We will put the origin of the y axi at the point where the key were thrown. Thi implifie thing in that the initial y coordinate of the key i y 0 = 0. Of coure, ince thi i a problem about free fall, we know the acceleration: a = g = 9.80 m 2. What mathematical information doe the problem give u? We are told that when t = 1.50, the y coordinate of the key i y = 4.00m. I thi enough information to olve the problem? We write the equation for y(t): y = y 0 + v 0 t at2 = v 0 t 1 2 gt2 where v 0 i preently unknown. At t = 1.50, y = 4.00m, o: 4.00m = v 0 (1.50) 1 2 (9.80 m 2 )(1.50) 2. Now we can olve for v 0. Rearrange thi equation to get: So: v 0 (1.50) = 4.00m (9.80 m 2 )(1.50) 2 = 15.0m. v 0 = 15.0m 1.50 = 10.0 m (b) We want to find the velocity of the key at the time they were caught, that i, at t = We know v 0 ; the velocity of the key at all time follow from Eq. 2.6, So at t = 1.50, v = v 0 + at = 10.0 m 9.80 m 2 t v = 10.0 m 9.80 m 2 (1.50) = 4.68 m. So the velocity of the key when they were caught wa 4.68 m. Note that the key had a negative velocity; thi tell u that the key were moving downward at the time they were caught! 16. A ball i thrown vertically upward from the ground with an initial peed of 15.0 m. (a) How long doe it take the ball to reach it maximum altitude? (b) What i it maximum altitude? (c) Determine the velocity and acceleration of the ball at t = [Ser4 2-49] (a) An illutration of the data given in thi problem i given in Fig We meaure the coordinate y upward from the place where the ball i thrown o that y 0 = 0. The ball acceleration while in flight i a = g = 9.80 m 2. We are given that v 0 = m. The ball i at maximum altitude when it (intantaneou) velocity v i zero (it i neither going up nor going down) and we can ue the expreion for v to olve for t: v = v 0 + at = t = v v 0 a

18 44 CHAPTER 2. MOTION IN ONE DIMENSION a = m/ 2 v = 0 m/ y v o = m/ Figure 2.9: Ball i thrown traight up with initial peed 15.0 m. Plug in the value for the top of the ball flight and get: t = (0) (15.0 m ) ( 9.80 m 2 ) = The ball take 1.53 to reach maximum height. (b) Now that we have the value of t when the ball i at maximum height we can plug it into Eq. 2.7 and find the value of y at thi time and that will be the value of the maximum height. But we can alo ue Eq. 2.8 ince we know all the value except for y. Solving for y we find: v 2 = v ay = y = v2 v 2 0 2a Plugging in the number, we get y = (0)2 (15.0 m )2 2( 9.80 m 2 ) = 11.5m The ball reache a maximum height of 11.5m. (c) At t = 2.00 (that i, 2.0 econd after the ball wa thrown) we ue Eq. 2.6 to find: v = v 0 + at = (15.0 m ) + ( 9.80 m 2 )(2.00) = 4.60 m. o at t = 2.00 the ball i on it way back down with a peed of 4.60 m. A for the next part, the acceleration of the ball i alway equal to 9.80 m 2 in flight. while it i 17. A baeball i hit uch that it travel traight upward after being truck by the bat. A fan oberve that it require 3.00 for the ball to reach it maximum height. Find (a) it initial velocity and (b) it maximum height. Ignore the effect of air reitance. [Ser4 2-51]

19 2.2. WORKED EXAMPLES 45 t = 3.00 v=0 v 0 Figure 2.10: Ball i hit traight up; reache maximum height 3.00 later. (a) An illutration of the data given in the problem i given in Fig For the period from when the ball i hit to the time it reache maximum height, we know the time interval, the acceleration (a = g) and alo the final velocity, ince at maximum height the velocity of the ball i zero. Then Eq. 2.6 give u v 0 : and we get: v = v 0 + at = v 0 = v at v 0 = 0 ( 9.80 m 2 )(3.00) = 29.4 m The initial velocity of the ball wa m. (b) To find the value of the maximum height, we need to find the value of the y coordinate at time t = We can ue either Eq. 2.7 or Eq the latter give: v 2 = v a(y y 0 ) = (y y 0 ) = v2 v 2 0 2a Plugging in the number we find that the change in y coordinate for the trip up wa: y y 0 = 02 (29.4 m )2 2( 9.80 m 2 ) = 44.1m. The ball reached a maximum height of 44.1m. 18. A parachutit bail out and freely fall 50 m. Then the parachute open, and thereafter he decelerate at 2.0 m 2. She reache the ground with a peed of 3.0 m. (a) How long wa the parachutit in the air? (b) At what height did the fall begin? [HRW5 2-84] (a) Thi problem give everal odd bit of information about the motion of the parachutit! We organize the information by drawing a diagram, like the one given in Fig It i

20 46 CHAPTER 2. MOTION IN ONE DIMENSION v=0 (a) Free Fall 50 m (b) Deceleration v=3.0 m/ (c) Figure 2.11: Diagram howing motion of parachutit in Example 18. very important to organize our work in thi way! At the height indicated by (a) in the figure, the kydiver ha zero initial peed. A he fall from (a) to (b) her acceleration i that of gravity, namely 9.80 m 2 downward. We know that (b) i 50 m lower than (a) but we don t yet know the kydiver peed at (b). Finally, at point (c) her peed i 3.0 m and between (b) and (c) her deceleration wa 2.0 m 2, but we don t know the difference in height between (b) and (c). How can we tart to fill in the gap in our knowledge? We note that on the trip from (a) to (b) we do know the tarting velocity, the ditance travelled and the acceleration. From Eq. 2.8 we can ee that thi i enough to find the final velocity, that i, the velocity at (b). Ue a coordinate ytem (y) which ha it origin at level (b), and the y axi pointing upward. Then the initial y coordinate i y 0 = 50m and the the initial velocity i v 0 = 0. The final y coordinate i y = 0 and the acceleration i contant at a = 9.80 m 2. Then uing Eq. 2.8 we have: v 2 = v a(y y + 0) = 0 + 2( 9.80 m 2 )(0 50m) = 980 m2 2 which ha the olution v = ±31.3 m but here the kydiver i obviouly moving downward at (b), o we mut pick v = 31.3 m for the velocity at (b). While we re at it, we can find the time it took to get from (a) to (b) uing Eq. 2.6, ince we know the velocitie and the acceleration for the motion. We find: v = v 0 + at = t = v v 0 a

21 2.2. WORKED EXAMPLES 47 Subtitute: t = ( 31.3 m 0) 9.80 m 2 = 3.19 The kydiver take 3.19 to fall from (a) to (b). Now we conider the motion from (b) to (c). For thi part of the motion we know the initial and final velocitie. We alo know the acceleration, but we mut be careful about how it i expreed. During thi part of the trip, the kydiver motion i alway downward (velocity i alway negative) but her peed decreae from 31.9 m to 3.0 m. The velocity change from 31.3 m to 3.0 m o that the velocity ha increaed. The acceleration i poitive; it i in the oppoite ene a the motion and thu it wa rightly called a deceleration in the problem. So for the motion from (b) to (c), we have a = +2.0 m 2 We have the tarting and final velocitie for the trip from (b) to (c) o Eq. 2.6 let u olve for the time t: v = v 0 + at = t = v v 0 a Subtitute: t = 3.0 m ( 31.3 m ) +2.0 m 2 = 14.2 Now we are prepared to anwer part (a) of the problem. The time of the travel from (a) wa 3.19; the time of travel from (b) to (c) wa The total time in the air wa t Total = = 17.4 (b) Let keep thinking about the trip from (b) to (c); we ll keep the origin at the ame place a before (at (b)). Then for the trip from (b) to (c) the initial coordinate i y 0 = 0. The initial velocity i v 0 = 31.9 m and the final velocity i v = 3.0 m. We have the acceleration, o Eq. 2.8 give u the final coordinate y: Subtitute: v 2 = v a(y y 0) = y y 0 = v2 v 2 0 2a y y 0 = ( 3.0 m )2 ( 31.3 m )2 2(+2.0 m 2 ) = 243m Since we choe y 0 = 0, the final coordinate of the kydiver i y = 243m. We have ued the ame coordinate ytem in both part, o overall the kydiver ha gone from y = +50m to y = 243m. The change in height wa y = 243m 50m = 293m So the parachutit fall began at a height of 293m above the ground.

22 48 CHAPTER 2. MOTION IN ONE DIMENSION 19. A tone fall from ret from the top of a high cliff. a econd tone i thrown downward from the ame height 2.00 later with an initial peed of 30.0 m. If both tone hit the ground imultaneouly, how high i the cliff? [Ser4 2-54] Thi i a puzzle type problem which goe beyond the normal ubtitute and olve type; it involve more organization of our work and a clear undertanding of our equation. Here the way I would attack it. We have two different falling object here with their own coordinate; we ll put our origin at the top of the cliff and call the y coordinate of the firt tone y 1 and that of the econd tone y 2. Each ha a different dependence on the time t. For the firt rock, we have v 0 = 0 ince it fall from ret and of coure a = g o that it poition i given by y 1 = y 0 + v 0 t at2 = 1 2 gt2 Thi i imple enough but we need to remind ourelve that here t i the time ince the firt tone tarted it motion. It i not the ame a the time ince the econd tone tart it motion. To be clear, let call thi time t 1. So we have: y 1 = 1 2 gt2 1 = 4.90 m 2 t 2 1 Now, for the motion of the econd tone, if we write t 2 for the time ince it tarted it motion, the fact tated in the problem tell u that it y coordinate i given by: y 2 = y 0 + v 0 t at2 2 = ( 30.0 m )t gt2 2 So far, o good. The problem tell u that the firt tone ha been falling for 2.0 longer than the econd one. Thi mean that t 1 i 2.0 larger than t 2. So: t 1 = t = t 2 = t (We will ue t 1 a our one time variable.) Putting thi into our lat equation and doing ome algebra give y 2 = ( 30.0 m )(t 1 2.0) 1 2 (9.80 m 2 )(t 1 2.0) 2 = ( 30.0 m )(t 1 2.0) (4.90 m 2 )(t t ) 2 = ( 4.90 m 2 )t ( 30.0 m m )t 1 + (60.0m 19.6m) = ( 4.90 m 2 )t ( 10.4 m )t 1 + (40.4m) We need to remember that thi expreion for y 2 will be meaningle for value of t 1 which are le than 2.0. With thi expreion we can find value of y 1 and y 2 uing the ame time coordinate, t 1. Now, the problem tell u that at ome time (t 1 ) the coordinate of the two tone are equal. We don t yet yet know what that time or coordinate i but that i the information contained in the tatement both tone hit the ground imultaneouly. We can find thi time by etting y 1 equal to y 2 and olving: ( 4.90 m 2 )t 2 1 = ( 4.90 m 2 )t ( 10.4 m )t 1 + (40.4m)

23 2.2. WORKED EXAMPLES 49 v A =0 A v B B 30.0 m 1.50 v C C Figure 2.12: Diagram for the falling object in Example 20. Fortunately the t 2 term cancel in thi equation making it a lot eaier. We get: ( 10.4 m )t 1 + (40.4m) = 0 which ha the olution t 1 = 40.4m 10.4 m = 3.88 So the rock will have the ame location at t 1 = 3.88, that i, 3.88 after the firt rock ha been dropped. What i that location? We can find thi by uing our value of t 1 to get either y 1 or y 2 (the anwer will be the ame). Putting it into the expreion for y 1 we get: y 1 = 4.90 m 2 t 2 1 = ( 4.90 m 2 )(3.88) 2 = 74m So both tone were 74m below the initial point at the time of impact. The cliff i high. 74m 20. A falling object require 1.50 to travel the lat 30.0m before hitting the ground. From what height above the ground did it fall? [Ser4 2-68] Thi i an intriguing ort of problem... very eay to tate, but not o clear a to where to begin in etting it up! The firt thing to do i draw a diagram. We draw the important point of the object motion, a in Fig The object ha zero velocity at A; at B it i at a height of 30.0m above the ground with an unknown velocity. At C it i at ground level, the time i 1.50 later than at B and we alo don t know the velocity here. Of coure, we know the acceleration: a = 9.80 m 2!! We are given all the information about the trip from B to C, o why not try to fill in our knowledge about thi part? We know the final and initial coordinate, the acceleration and the time o we can find the initial velocity (that i, the velocity at B). Let put the origin at ground level; then, y 0 = 30.0m, y = 0 and t = 1.50, and uing y = y 0 + v 0 t at2

24 50 CHAPTER 2. MOTION IN ONE DIMENSION we find: o that v 0 t = (y y 0 ) 1 2 at2 = (0 (30.0m)) 1 2 ( 9.80 m 2 )(1.50) 2 = 19.0m v 0 = ( 19.0m) t = ( 19.0m) (1.50 ) = 12.5 m. Thi i the velocity at point B; we can alo find the velocity at C eaily, ince that i the final velocity, v: v = v 0 + at = ( 12.5 m ) + ( 9.80 m 2 )(1.50) = 27.3 m Now we can conider the trip from the tarting point, A to the point of impact, C. We don t know the initial y coordinate, but we do know the final and initial velocitie: The initial velocity i v 0 = 0 and the final velocity i v = 27.3 m, a we jut found. With the origin et at ground level, the final y coordinate i y = 0. We don t know the time for the trip, but if we ue: v 2 = v a(y y 0 ) we find: (y y 0 ) = (v2 v 2 0) 2a and we can rearrange thi to get: = ( 27.3 m )2 (0) 2 2( 9.80 m 2 ) y 0 = y m = m = 38.2m = 38.2m and the o the object tarted falling from a height of 38.2m. There are probably cleverer way to do thi problem, but here I wanted to give you the low, patient approach! 21. A tudent i taring idly out her dormitory window when he ee a water balloon fall pat. If the balloon take 0.22 to cro the 130cm high window, from what height above the top of the window wa it dropped? [Wolf 2-78] I will et up the vertical coordinate y a hown in Fig The origin i at the place where the balloon wa dropped, and we don t know how far above the window that i. Note, the y axi point downward here, o that y a a function of time i given by y = 1 2 gt2. Uing thi ytem, yet the y coordinate of the top of the window bet y 1 and the bottom of the window be y 2. Suppoe the balloon croe the top of the window at t 1 and the bottom of the window at t 2. The problem tell u that y 2 y 1 = 1.30m and t 2 t 1 = 0.22 Uing the equation of motion for the balloon, we have y 1 = 1 2 gt2 1 and y 2 = 1 2 gt2 2

25 2.2. WORKED EXAMPLES 51 0 y y 1, t 1 y 2, t 2 Figure 2.13: Diagram for the falling object in Example 21. In fact at thi point the problem i really olved becaue we have four equation for the four unknown y 1, y 2, t 1 and t 2. We jut need to do ome math! One way to olve the equation i to ubtitute for the y a: y 2 y 1 = 1 2 gt gt2 1 = 1 2 g(t2 2 t 2 1) = 1.30m But here we can factor the term t 2 2 t 2 1 to give: Thi give u t 2 + t 1 : t 2 + t 1 = 1 2 g(t2 2 t 2 1) = 1 2 g(t 2 + t 1 )(t 2 t 1 ) = 1.30m Adding thi to the equation t 2 t 1 = 0.22 give 2(1.30 m) (9.80 m )(t 2 2 t 1 ) = 2(1.30m) (9.80 m )(0.22) = t 2 = 1.43 = t 2 = 0.71 = t 1 = 0.49 And then the equation for y 1 give u y 1 = 1 2 gt2 1 = 1 2 (9.80 m 2 )(0.492) 2 = 1.19m o that the balloon began it fall 1.19m above the top of the window.

26 52 CHAPTER 2. MOTION IN ONE DIMENSION

MECH 2110 - Statics & Dynamics

MECH 2110 - Statics & Dynamics Chapter D Problem 3 Solution 1/7/8 1:8 PM MECH 11 - Static & Dynamic Chapter D Problem 3 Solution Page 7, Engineering Mechanic - Dynamic, 4th Edition, Meriam and Kraige Given: Particle moving along a traight

More information

Linear Momentum and Collisions

Linear Momentum and Collisions Chapter 7 Linear Momentum and Colliion 7.1 The Important Stuff 7.1.1 Linear Momentum The linear momentum of a particle with ma m moving with velocity v i defined a p = mv (7.1) Linear momentum i a vector.

More information

Unit 11 Using Linear Regression to Describe Relationships

Unit 11 Using Linear Regression to Describe Relationships Unit 11 Uing Linear Regreion to Decribe Relationhip Objective: To obtain and interpret the lope and intercept of the leat quare line for predicting a quantitative repone variable from a quantitative explanatory

More information

Discussion Session 4 Projectile Motion Week 05. The Plan

Discussion Session 4 Projectile Motion Week 05. The Plan PHYS Dicuion Seion 4 Projectile Motion Week 5 The Plan Thi week your group will practice analyzing projectile otion ituation. Why do we pend a whole eion on thi topic? The anwer i that projectile otion

More information

Physics 111. Exam #1. January 24, 2014

Physics 111. Exam #1. January 24, 2014 Phyic 111 Exam #1 January 24, 2014 Name Pleae read and follow thee intruction carefully: Read all problem carefully before attempting to olve them. Your work mut be legible, and the organization clear.

More information

Description: Conceptual questions about projectile motion and some easy calculations. (uses applets)

Description: Conceptual questions about projectile motion and some easy calculations. (uses applets) Week 3: Chapter 3 [ Edit ] Overview Suary View Diagnotic View Print View with Anwer Week 3: Chapter 3 Due: 11:59p on Sunday, February 8, 2015 To undertand how point are awarded, read the Grading Policy

More information

Newton s Laws. A force is simply a push or a pull. Forces are vectors; they have both size and direction.

Newton s Laws. A force is simply a push or a pull. Forces are vectors; they have both size and direction. Newton Law Newton firt law: An object will tay at ret or in a tate of uniform motion with contant velocity, in a traight line, unle acted upon by an external force. In other word, the bodie reit any change

More information

MSc Financial Economics: International Finance. Bubbles in the Foreign Exchange Market. Anne Sibert. Revised Spring 2013. Contents

MSc Financial Economics: International Finance. Bubbles in the Foreign Exchange Market. Anne Sibert. Revised Spring 2013. Contents MSc Financial Economic: International Finance Bubble in the Foreign Exchange Market Anne Sibert Revied Spring 203 Content Introduction................................................. 2 The Mone Market.............................................

More information

A) When two objects slide against one another, the magnitude of the frictional force is always equal to μ

A) When two objects slide against one another, the magnitude of the frictional force is always equal to μ Phyic 100 Homewor 5 Chapter 6 Contact Force Introduced ) When two object lide againt one another, the magnitude of the frictional force i alway equal to μ B) When two object are in contact with no relative

More information

Harmonic Oscillations / Complex Numbers

Harmonic Oscillations / Complex Numbers Harmonic Ocillation / Complex Number Overview and Motivation: Probably the ingle mot important problem in all of phyic i the imple harmonic ocillator. It can be tudied claically or uantum mechanically,

More information

Projectile Motion. Vectors and Projectiles

Projectile Motion. Vectors and Projectiles Nae: Projectile Motion Read fro Leon 2 of the Vector and Motion in Two-Dienion chapter at The Phyic Claroo: http://www.phyicclaroo.co/cla/vector/u3l2a.htl http://www.phyicclaroo.co/cla/vector/u3l2b.htl

More information

Catapults and Projectile Motion Parabolic Equations

Catapults and Projectile Motion Parabolic Equations Catapult and Projectile Motion Parabolic Equation OBJECTIVES Student will ue catapult to eplore projectile motion and parabolic equation. Student will erif the kinematic equation uing real-life data. MATERIALS

More information

Chapter 10 Velocity, Acceleration, and Calculus

Chapter 10 Velocity, Acceleration, and Calculus Chapter 10 Velocity, Acceleration, and Calculu The firt derivative of poition i velocity, and the econd derivative i acceleration. Thee derivative can be viewed in four way: phyically, numerically, ymbolically,

More information

Quadrilaterals. Learning Objectives. Pre-Activity

Quadrilaterals. Learning Objectives. Pre-Activity Section 3.4 Pre-Activity Preparation Quadrilateral Intereting geometric hape and pattern are all around u when we tart looking for them. Examine a row of fencing or the tiling deign at the wimming pool.

More information

Solution. Section 2.8 Related Rates. Exercise. Exercise. Exercise. dx when y 2. and. 3, then what is when x 1. Solution dy dy.

Solution. Section 2.8 Related Rates. Exercise. Exercise. Exercise. dx when y 2. and. 3, then what is when x 1. Solution dy dy. Section.8 Related Rate Eercie If y and d, then what i when 1 d d 6 6 1 1 6 Eercie If y y and 5, then what i d when y d d y 1 5 5 y 1 d 5 1 55 y Eercie A cube urface area increae at the rate of 7 when the

More information

Rotation of an Object About a Fixed Axis

Rotation of an Object About a Fixed Axis Chapter 1 Rotation of an Object About a Fixed Axi 1.1 The Important Stuff 1.1.1 Rigid Bodie; Rotation So far in our tudy of phyic we have (with few exception) dealt with particle, object whoe patial dimenion

More information

Chapter 32. OPTICAL IMAGES 32.1 Mirrors

Chapter 32. OPTICAL IMAGES 32.1 Mirrors Chapter 32 OPTICAL IMAGES 32.1 Mirror The point P i called the image or the virtual image of P (light doe not emanate from it) The left-right reveral in the mirror i alo called the depth inverion (the

More information

6. Friction, Experiment and Theory

6. Friction, Experiment and Theory 6. Friction, Experiment and Theory The lab thi wee invetigate the rictional orce and the phyical interpretation o the coeicient o riction. We will mae ue o the concept o the orce o gravity, the normal

More information

4.1 Radian and Degree Measure

4.1 Radian and Degree Measure 4. Radian and Degree Meaure An angle AOB (notation: AOB ) conit of two ray R and R with a common vertex O (ee Figure below). We often interpret an angle a a rotation of the ray R onto R. In thi cae, R

More information

2After completing this chapter you should be able to

2After completing this chapter you should be able to After completing this chapter you should be able to solve problems involving motion in a straight line with constant acceleration model an object moving vertically under gravity understand distance time

More information

Two Dimensional FEM Simulation of Ultrasonic Wave Propagation in Isotropic Solid Media using COMSOL

Two Dimensional FEM Simulation of Ultrasonic Wave Propagation in Isotropic Solid Media using COMSOL Excerpt from the Proceeding of the COMSO Conference 0 India Two Dimenional FEM Simulation of Ultraonic Wave Propagation in Iotropic Solid Media uing COMSO Bikah Ghoe *, Krihnan Balaubramaniam *, C V Krihnamurthy

More information

Solution of the Heat Equation for transient conduction by LaPlace Transform

Solution of the Heat Equation for transient conduction by LaPlace Transform Solution of the Heat Equation for tranient conduction by LaPlace Tranform Thi notebook ha been written in Mathematica by Mark J. McCready Profeor and Chair of Chemical Engineering Univerity of Notre Dame

More information

Name: SID: Instructions

Name: SID: Instructions CS168 Fall 2014 Homework 1 Aigned: Wedneday, 10 September 2014 Due: Monday, 22 September 2014 Name: SID: Dicuion Section (Day/Time): Intruction - Submit thi homework uing Pandagrader/GradeScope(http://www.gradecope.com/

More information

Physics Kinematics Model

Physics Kinematics Model Physics Kinematics Model I. Overview Active Physics introduces the concept of average velocity and average acceleration. This unit supplements Active Physics by addressing the concept of instantaneous

More information

Projectile motion simulator. http://www.walter-fendt.de/ph11e/projectile.htm

Projectile motion simulator. http://www.walter-fendt.de/ph11e/projectile.htm More Chapter 3 Projectile motion simulator http://www.walter-fendt.de/ph11e/projectile.htm The equations of motion for constant acceleration from chapter 2 are valid separately for both motion in the x

More information

You may use a scientific calculator (non-graphing, non-programmable) during testing.

You may use a scientific calculator (non-graphing, non-programmable) during testing. TECEP Tet Decription College Algebra MAT--TE Thi TECEP tet algebraic concept, procee, and practical application. Topic include: linear equation and inequalitie; quadratic equation; ytem of equation and

More information

Modelling and Solving Two-Step Equations: a(x + b) = c

Modelling and Solving Two-Step Equations: a(x + b) = c Modelling and Solving Two-Step Equation: a( + b) c Focu on After thi leon, you will be able to model problem with two-tep linear equation olve two-tep linear equation and how how you worked out the anwer

More information

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science aachuett Intitute of Technology Department of Electrical Engineering and Computer Science 6.685 Electric achinery Cla Note 10: Induction achine Control and Simulation c 2003 Jame L. Kirtley Jr. 1 Introduction

More information

A technical guide to 2014 key stage 2 to key stage 4 value added measures

A technical guide to 2014 key stage 2 to key stage 4 value added measures A technical guide to 2014 key tage 2 to key tage 4 value added meaure CONTENTS Introduction: PAGE NO. What i value added? 2 Change to value added methodology in 2014 4 Interpretation: Interpreting chool

More information

FLUID MECHANICS. TUTORIAL No.4 FLOW THROUGH POROUS PASSAGES

FLUID MECHANICS. TUTORIAL No.4 FLOW THROUGH POROUS PASSAGES FLUID MECHANICS TUTORIAL No.4 FLOW THROUGH POROUS PASSAGES In thi tutorial you will continue the work on laminar flow and develop Poieuille' equation to the form known a the Carman - Kozeny equation. Thi

More information

Queueing systems with scheduled arrivals, i.e., appointment systems, are typical for frontal service systems,

Queueing systems with scheduled arrivals, i.e., appointment systems, are typical for frontal service systems, MANAGEMENT SCIENCE Vol. 54, No. 3, March 28, pp. 565 572 in 25-199 ein 1526-551 8 543 565 inform doi 1.1287/mnc.17.82 28 INFORMS Scheduling Arrival to Queue: A Single-Server Model with No-Show INFORMS

More information

Simple Harmonic Motion. AP Physics B

Simple Harmonic Motion. AP Physics B Simple Harmonic Motion AP Phyic B Simple Harmonic Motion Back and forth motion that i caued by a force that i directly proportional to the diplacement. The diplacement center around an equilibrium poition.

More information

Exam 1 Review Questions PHY 2425 - Exam 1

Exam 1 Review Questions PHY 2425 - Exam 1 Exam 1 Review Questions PHY 2425 - Exam 1 Exam 1H Rev Ques.doc - 1 - Section: 1 7 Topic: General Properties of Vectors Type: Conceptual 1 Given vector A, the vector 3 A A) has a magnitude 3 times that

More information

Chapter 3 Solutions. Figure 3.7a. (b) Thus (c) velocity: At. Figure 3.7b

Chapter 3 Solutions. Figure 3.7a. (b) Thus (c) velocity: At. Figure 3.7b Chapter 3 Solutions 3.7.IDENTIFY and Use Eqs. (3.4) and (3.12) to find and as functions of time. The magnitude and direction of and can be found once we know their components. (a) Calculate x and y for

More information

Incline and Friction Examples

Incline and Friction Examples Incline and riction Eample Phic 6A Prepared b Vince Zaccone riction i a force that oppoe the motion of urface that are in contact with each other. We will conider 2 tpe of friction in thi cla: KINETIC

More information

Chapter 4: Mean-Variance Analysis

Chapter 4: Mean-Variance Analysis Chapter 4: Mean-Variance Analyi Modern portfolio theory identifie two apect of the invetment problem. Firt, an invetor will want to maximize the expected rate of return on the portfolio. Second, an invetor

More information

1 of 7 9/5/2009 6:12 PM

1 of 7 9/5/2009 6:12 PM 1 of 7 9/5/2009 6:12 PM Chapter 2 Homework Due: 9:00am on Tuesday, September 8, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]

More information

Original Article: TOWARDS FLUID DYNAMICS EQUATIONS

Original Article: TOWARDS FLUID DYNAMICS EQUATIONS Peer Reviewed, Open Acce, Free Online Journal Publihed monthly : ISSN: 8-8X Iue 4(5); April 15 Original Article: TOWARDS FLUID DYNAMICS EQUATIONS Citation Zaytev M.L., Akkerman V.B., Toward Fluid Dynamic

More information

Introduction to the article Degrees of Freedom.

Introduction to the article Degrees of Freedom. Introduction to the article Degree of Freedom. The article by Walker, H. W. Degree of Freedom. Journal of Educational Pychology. 3(4) (940) 53-69, wa trancribed from the original by Chri Olen, George Wahington

More information

Initial & Final Value Theorems. Lecture 7. More on Laplace Transform (Lathi ) Example. Laplace Transform for Solving Differential Equations

Initial & Final Value Theorems. Lecture 7. More on Laplace Transform (Lathi ) Example. Laplace Transform for Solving Differential Equations Initial & Final Value Theorem ecture 7 More on aplace Tranform (athi 4.3 4.4) How to find the initial and final value of a function x(t) if we now it aplace Tranform? (t 0 +, and t ) Initial Value Theorem

More information

Physics Exam 1 Review - Chapter 1,2

Physics Exam 1 Review - Chapter 1,2 Physics 1401 - Exam 1 Review - Chapter 1,2 13. Which of the following is NOT one of the fundamental units in the SI system? A) newton B) meter C) kilogram D) second E) All of the above are fundamental

More information

Scalar versus Vector Quantities. Speed. Speed: Example Two. Scalar Quantities. Average Speed = distance (in meters) time (in seconds) v =

Scalar versus Vector Quantities. Speed. Speed: Example Two. Scalar Quantities. Average Speed = distance (in meters) time (in seconds) v = Scalar versus Vector Quantities Scalar Quantities Magnitude (size) 55 mph Speed Average Speed = distance (in meters) time (in seconds) Vector Quantities Magnitude (size) Direction 55 mph, North v = Dx

More information

Ohm s Law. Ohmic relationship V=IR. Electric Power. Non Ohmic devises. Schematic representation. Electric Power

Ohm s Law. Ohmic relationship V=IR. Electric Power. Non Ohmic devises. Schematic representation. Electric Power Ohm Law Ohmic relationhip V=IR Ohm law tate that current through the conductor i directly proportional to the voltage acro it if temperature and other phyical condition do not change. In many material,

More information

12.4 Problems. Excerpt from "Introduction to Geometry" 2014 AoPS Inc. Copyrighted Material CHAPTER 12. CIRCLES AND ANGLES

12.4 Problems. Excerpt from Introduction to Geometry 2014 AoPS Inc.  Copyrighted Material CHAPTER 12. CIRCLES AND ANGLES HTER 1. IRLES N NGLES Excerpt from "Introduction to Geometry" 014 os Inc. onider the circle with diameter O. all thi circle. Why mut hit O in at leat two di erent point? (b) Why i it impoible for to hit

More information

The Nonlinear Pendulum

The Nonlinear Pendulum The Nonlinear Pendulum D.G. Simpon, Ph.D. Department of Phyical Science and Enineerin Prince Geore ommunity ollee December 31, 1 1 The Simple Plane Pendulum A imple plane pendulum conit, ideally, of a

More information

Senior Thesis. Horse Play. Optimal Wagers and the Kelly Criterion. Author: Courtney Kempton. Supervisor: Professor Jim Morrow

Senior Thesis. Horse Play. Optimal Wagers and the Kelly Criterion. Author: Courtney Kempton. Supervisor: Professor Jim Morrow Senior Thei Hore Play Optimal Wager and the Kelly Criterion Author: Courtney Kempton Supervior: Profeor Jim Morrow June 7, 20 Introduction The fundamental problem in gambling i to find betting opportunitie

More information

Three Phase Theory - Professor J R Lucas

Three Phase Theory - Professor J R Lucas Three Phae Theory - Profeor J Luca A you are aware, to tranit power with ingle phae alternating current, we need two wire live wire and neutral. However you would have een that ditribution line uually

More information

Chapter H - Problems

Chapter H - Problems Chapter H - Problem Blinn College - Phyic 45 - Terry Honan Problem H.1 A wheel rotate from ret to 1 ê in 3. Aume the angular acceleration i contant. (a) What i the magnitude of the wheel' angular acceleration?

More information

Chapter 10 Stocks and Their Valuation ANSWERS TO END-OF-CHAPTER QUESTIONS

Chapter 10 Stocks and Their Valuation ANSWERS TO END-OF-CHAPTER QUESTIONS Chapter Stoc and Their Valuation ANSWERS TO EN-OF-CHAPTER QUESTIONS - a. A proxy i a document giving one peron the authority to act for another, typically the power to vote hare of common toc. If earning

More information

Math 22B, Homework #8 1. y 5y + 6y = 2e t

Math 22B, Homework #8 1. y 5y + 6y = 2e t Math 22B, Homework #8 3.7 Problem # We find a particular olution of the ODE y 5y + 6y 2e t uing the method of variation of parameter and then verify the olution uing the method of undetermined coefficient.

More information

ECE 320 Energy Conversion and Power Electronics Dr. Tim Hogan. Chapter 7: Synchronous Machines and Drives (Textbook Chapter 5)

ECE 320 Energy Conversion and Power Electronics Dr. Tim Hogan. Chapter 7: Synchronous Machines and Drives (Textbook Chapter 5) ECE 30 Energy Converion and Power Electronic Dr. Tim Hogan Chapter 7: ynchronou Machine and Drive (Textbook Chapter 5) Chapter Objective For induction machine, a the rotor approache ynchronou peed, the

More information

Phys 201 Fall 2009 Thursday, September 17, 2009 & Tuesday, September 19, Chapter 3: Mo?on in Two and Three Dimensions

Phys 201 Fall 2009 Thursday, September 17, 2009 & Tuesday, September 19, Chapter 3: Mo?on in Two and Three Dimensions Phys 201 Fall 2009 Thursday, September 17, 2009 & Tuesday, September 19, 2009 Chapter 3: Mo?on in Two and Three Dimensions Displacement, Velocity and Acceleration Displacement describes the location of

More information

1D STEADY STATE HEAT

1D STEADY STATE HEAT D SEADY SAE HEA CONDUCION () Prabal alukdar Aociate Profeor Department of Mechanical Engineering II Delhi E-mail: prabal@mech.iitd.ac.in Convection Boundary Condition Heat conduction at the urface in a

More information

People s Physics book

People s Physics book The Big Idea In this chapter, we aim to understand and explain the parabolic motion of a thrown object, known as projectile motion. Motion in one direction is unrelated to motion in other perpendicular

More information

with "a", "b" and "c" representing real numbers, and "a" is not equal to zero.

with a, b and c representing real numbers, and a is not equal to zero. 3.1 SOLVING QUADRATIC EQUATIONS: * A QUADRATIC is a polynomial whose highest exponent is. * The "standard form" of a quadratic equation is: ax + bx + c = 0 with "a", "b" and "c" representing real numbers,

More information

On Reference RIAA Networks by Jim Hagerman

On Reference RIAA Networks by Jim Hagerman On eference IAA Network by Jim Hagerman You d think there would be nothing left to ay. Everything you need to know about IAA network ha already been publihed. However, a few year back I came acro an intereting

More information

Engineering Bernoulli Equation

Engineering Bernoulli Equation Engineering Bernoulli Equation R. Shankar Subramanian Department of Chemical and Biomolecular Engineering Clarkon Univerity The Engineering Bernoulli equation can be derived from the principle of conervation

More information

A note on profit maximization and monotonicity for inbound call centers

A note on profit maximization and monotonicity for inbound call centers A note on profit maximization and monotonicity for inbound call center Ger Koole & Aue Pot Department of Mathematic, Vrije Univeriteit Amterdam, The Netherland 23rd December 2005 Abtract We conider an

More information

Block Diagrams, State-Variable Models, and Simulation Methods

Block Diagrams, State-Variable Models, and Simulation Methods 5 C H A P T E R Block Diagram, State-Variable Model, and Simulation Method CHAPTER OUTLINE CHAPTER OBJECTIVES Part I. Model Form 25 5. Tranfer Function and Block Diagram Model 25 5.2 State-Variable Model

More information

Report 4668-1b 30.10.2010. Measurement report. Sylomer - field test

Report 4668-1b 30.10.2010. Measurement report. Sylomer - field test Report 4668-1b Meaurement report Sylomer - field tet Report 4668-1b 2(16) Contet 1 Introduction... 3 1.1 Cutomer... 3 1.2 The ite and purpoe of the meaurement... 3 2 Meaurement... 6 2.1 Attenuation of

More information

Pipe Flow Calculations

Pipe Flow Calculations Pipe Flow Calculation R. Shankar Subramanian epartment o Chemical and Biomolecular Engineering Clarkon Univerity We begin with ome reult that we hall ue when making riction lo calculation or teady, ully

More information

Time hours. 1. Above is a velocity time graph of a moving car. Answer the following questions using the graph. a. At what time was the car stopped?

Time hours. 1. Above is a velocity time graph of a moving car. Answer the following questions using the graph. a. At what time was the car stopped? Time hours 1. Above is a velocity time graph of a moving car. Answer the following questions using the graph. a. At what time was the car stopped? b. At what time did the car have the greatest velocity?

More information

σ m using Equation 8.1 given that σ

σ m using Equation 8.1 given that σ 8. Etimate the theoretical fracture trength of a brittle material if it i known that fracture occur by the propagation of an elliptically haped urface crack of length 0.8 mm and having a tip radiu of curvature

More information

Assessing the Discriminatory Power of Credit Scores

Assessing the Discriminatory Power of Credit Scores Aeing the Dicriminatory Power of Credit Score Holger Kraft 1, Gerald Kroiandt 1, Marlene Müller 1,2 1 Fraunhofer Intitut für Techno- und Wirtchaftmathematik (ITWM) Gottlieb-Daimler-Str. 49, 67663 Kaierlautern,

More information

Chapter 4. Motion in two & three dimensions

Chapter 4. Motion in two & three dimensions Chapter 4 Motion in two & three dimensions 4.2 Position and Displacement Position The position of a particle can be described by a position vector, with respect to a reference origin. Displacement The

More information

f max s = µ s N (5.1)

f max s = µ s N (5.1) Chapter 5 Forces and Motion II 5.1 The Important Stuff 5.1.1 Friction Forces Forces which are known collectively as friction forces are all around us in daily life. In elementary physics we discuss the

More information

2) When you look at the speedometer in a moving car, you can see the car's.

2) When you look at the speedometer in a moving car, you can see the car's. Practice Kinematics Questions Answers are at the end Choose the best answer to each question and write the appropriate letter in the space provided. 1) One possible unit of speed is. A) light years per

More information

DISTRIBUTED DATA PARALLEL TECHNIQUES FOR CONTENT-MATCHING INTRUSION DETECTION SYSTEMS. G. Chapman J. Cleese E. Idle

DISTRIBUTED DATA PARALLEL TECHNIQUES FOR CONTENT-MATCHING INTRUSION DETECTION SYSTEMS. G. Chapman J. Cleese E. Idle DISTRIBUTED DATA PARALLEL TECHNIQUES FOR CONTENT-MATCHING INTRUSION DETECTION SYSTEMS G. Chapman J. Cleee E. Idle ABSTRACT Content matching i a neceary component of any ignature-baed network Intruion Detection

More information

Optical Illusion. Sara Bolouki, Roger Grosse, Honglak Lee, Andrew Ng

Optical Illusion. Sara Bolouki, Roger Grosse, Honglak Lee, Andrew Ng Optical Illuion Sara Bolouki, Roger Groe, Honglak Lee, Andrew Ng. Introduction The goal of thi proect i to explain ome of the illuory phenomena uing pare coding and whitening model. Intead of the pare

More information

8. As a cart travels around a horizontal circular track, the cart must undergo a change in (1) velocity (3) speed (2) inertia (4) weight

8. As a cart travels around a horizontal circular track, the cart must undergo a change in (1) velocity (3) speed (2) inertia (4) weight 1. What is the average speed of an object that travels 6.00 meters north in 2.00 seconds and then travels 3.00 meters east in 1.00 second? 9.00 m/s 3.00 m/s 0.333 m/s 4.24 m/s 2. What is the distance traveled

More information

Review of Multiple Regression Richard Williams, University of Notre Dame, http://www3.nd.edu/~rwilliam/ Last revised January 13, 2015

Review of Multiple Regression Richard Williams, University of Notre Dame, http://www3.nd.edu/~rwilliam/ Last revised January 13, 2015 Review of Multiple Regreion Richard William, Univerity of Notre Dame, http://www3.nd.edu/~rwilliam/ Lat revied January 13, 015 Aumption about prior nowledge. Thi handout attempt to ummarize and yntheize

More information

2. METHOD DATA COLLECTION

2. METHOD DATA COLLECTION Key to learning in pecific ubject area of engineering education an example from electrical engineering Anna-Karin Cartenen,, and Jonte Bernhard, School of Engineering, Jönköping Univerity, S- Jönköping,

More information

Project Management Basics

Project Management Basics Project Management Baic A Guide to undertanding the baic component of effective project management and the key to ucce 1 Content 1.0 Who hould read thi Guide... 3 1.1 Overview... 3 1.2 Project Management

More information

Thus far. Inferences When Comparing Two Means. Testing differences between two means or proportions

Thus far. Inferences When Comparing Two Means. Testing differences between two means or proportions Inference When Comparing Two Mean Dr. Tom Ilvento FREC 48 Thu far We have made an inference from a ingle ample mean and proportion to a population, uing The ample mean (or proportion) The ample tandard

More information

1) Assume that the sample is an SRS. The problem state that the subjects were randomly selected.

1) Assume that the sample is an SRS. The problem state that the subjects were randomly selected. 12.1 Homework for t Hypothei Tet 1) Below are the etimate of the daily intake of calcium in milligram for 38 randomly elected women between the age of 18 and 24 year who agreed to participate in a tudy

More information

Section 2.2 Arc Length and Sector Area. Arc Length. Definition. Note:

Section 2.2 Arc Length and Sector Area. Arc Length. Definition. Note: Section. Arc Length and Sector Area Arc Length Definition If a central angle, in a circle of a radiu r, cut off an arc of length, then the meaure of, in radian i: r r r r ( in radian) Note: When applying

More information

DISTRIBUTED DATA PARALLEL TECHNIQUES FOR CONTENT-MATCHING INTRUSION DETECTION SYSTEMS

DISTRIBUTED DATA PARALLEL TECHNIQUES FOR CONTENT-MATCHING INTRUSION DETECTION SYSTEMS DISTRIBUTED DATA PARALLEL TECHNIQUES FOR CONTENT-MATCHING INTRUSION DETECTION SYSTEMS Chritopher V. Kopek Department of Computer Science Wake Foret Univerity Winton-Salem, NC, 2709 Email: kopekcv@gmail.com

More information

Review Chapters 2, 3, 4, 5

Review Chapters 2, 3, 4, 5 Review Chapters 2, 3, 4, 5 4) The gain in speed each second for a freely-falling object is about A) 0. B) 5 m/s. C) 10 m/s. D) 20 m/s. E) depends on the initial speed 9) Whirl a rock at the end of a string

More information

1 Safe Drivers versus Reckless Drunk Drivers

1 Safe Drivers versus Reckless Drunk Drivers ECON 301: General Equilibrium IV (Externalitie) 1 Intermediate Microeconomic II, ECON 301 General Equilibrium IV: Externalitie In our dicuion thu far, we have implicitly aumed that all good can be traded

More information

IMPORTANT: Read page 2 ASAP. *Please feel free to email (longo.physics@gmail.com) me at any time if you have questions or concerns.

IMPORTANT: Read page 2 ASAP. *Please feel free to email (longo.physics@gmail.com) me at any time if you have questions or concerns. rev. 05/4/16 AP Phyic C: Mechanic Summer Aignment 016-017 Mr. Longo Foret Park HS longo.phyic@gmail.com longodb@pwc.edu Welcome to AP Phyic C: Mechanic. The purpoe of thi ummer aignment i to give you a

More information

A Note on Profit Maximization and Monotonicity for Inbound Call Centers

A Note on Profit Maximization and Monotonicity for Inbound Call Centers OPERATIONS RESEARCH Vol. 59, No. 5, September October 2011, pp. 1304 1308 in 0030-364X ein 1526-5463 11 5905 1304 http://dx.doi.org/10.1287/opre.1110.0990 2011 INFORMS TECHNICAL NOTE INFORMS hold copyright

More information

In order to describe motion you need to describe the following properties.

In order to describe motion you need to describe the following properties. Chapter 2 One Dimensional Kinematics How would you describe the following motion? Ex: random 1-D path speeding up and slowing down In order to describe motion you need to describe the following properties.

More information

HOMOTOPY PERTURBATION METHOD FOR SOLVING A MODEL FOR HIV INFECTION OF CD4 + T CELLS

HOMOTOPY PERTURBATION METHOD FOR SOLVING A MODEL FOR HIV INFECTION OF CD4 + T CELLS İtanbul icaret Üniveritei Fen Bilimleri Dergii Yıl: 6 Sayı: Güz 7/. 9-5 HOMOOPY PERURBAION MEHOD FOR SOLVING A MODEL FOR HIV INFECION OF CD4 + CELLS Mehmet MERDAN ABSRAC In thi article, homotopy perturbation

More information

The Magic Chart Honors Physics

The Magic Chart Honors Physics The Magic Chart Honors Physics Magic Chart Equations v = v o + a t x = v o t + 1/2 a t 2 x = ½ (v o + v) t v 2 = v 2 o + 2a x x = vt - 1/2 a t 2 x Who Cares Quantity v a t v o THE WHO CARES QUANTITY tells

More information

Module 8. Three-phase Induction Motor. Version 2 EE IIT, Kharagpur

Module 8. Three-phase Induction Motor. Version 2 EE IIT, Kharagpur Module 8 Three-phae Induction Motor Verion EE IIT, Kharagpur Leon 33 Different Type of Starter for Induction Motor (IM Verion EE IIT, Kharagpur Inructional Objective Need of uing arter for Induction motor

More information

Chapter 4. Kinematics - Velocity and Acceleration. 4.1 Purpose. 4.2 Introduction

Chapter 4. Kinematics - Velocity and Acceleration. 4.1 Purpose. 4.2 Introduction Chapter 4 Kinematics - Velocity and Acceleration 4.1 Purpose In this lab, the relationship between position, velocity and acceleration will be explored. In this experiment, friction will be neglected.

More information

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. Dr Tay Seng Chuan

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. Dr Tay Seng Chuan Ground Rules PC11 Fundamentals of Physics I Lectures 3 and 4 Motion in One Dimension Dr Tay Seng Chuan 1 Switch off your handphone and pager Switch off your laptop computer and keep it No talking while

More information

DIHEDRAL GROUPS KEITH CONRAD

DIHEDRAL GROUPS KEITH CONRAD DIHEDRAL GROUPS KEITH CONRAD 1. Introduction For n 3, the dihedral group D n i defined a the rigid motion 1 of the plane preerving a regular n-gon, with the operation being compoition. Thee polygon for

More information

Bio-Plex Analysis Software

Bio-Plex Analysis Software Multiplex Supenion Array Bio-Plex Analyi Software The Leader in Multiplex Immunoaay Analyi Bio-Plex Analyi Software If making ene of your multiplex data i your challenge, then Bio-Plex data analyi oftware

More information

Physics: Principles and Applications, 6e Giancoli Chapter 2 Describing Motion: Kinematics in One Dimension

Physics: Principles and Applications, 6e Giancoli Chapter 2 Describing Motion: Kinematics in One Dimension Physics: Principles and Applications, 6e Giancoli Chapter 2 Describing Motion: Kinematics in One Dimension Conceptual Questions 1) Suppose that an object travels from one point in space to another. Make

More information

The quest to find how x(t) and y(t) depend on t is greatly simplified by the following facts, first discovered by Galileo:

The quest to find how x(t) and y(t) depend on t is greatly simplified by the following facts, first discovered by Galileo: Team: Projectile Motion So far you have focused on motion in one dimension: x(t). In this lab, you will study motion in two dimensions: x(t), y(t). This 2D motion, called projectile motion, consists of

More information

Latitude dependence of the maximum duration of a total solar eclipse

Latitude dependence of the maximum duration of a total solar eclipse Latitue epenence of the axiu uration of a total olar eclipe Author: Jen Buu, with aitance fro Jean Meeu Contact: 6 Baker Street, Gayton, Northant, NN7 3EZ, UK jbuu@btinternet.co Introuction It i well known

More information

Finite Automata. a) Reading a symbol, b) Transferring to a new instruction, and c) Advancing the tape head one square to the right.

Finite Automata. a) Reading a symbol, b) Transferring to a new instruction, and c) Advancing the tape head one square to the right. Finite Automata Let u begin by removing almot all of the Turing machine' power! Maybe then we hall have olvable deciion problem and till be able to accomplih ome computational tak. Alo, we might be able

More information

SOLUTIONS TO CONCEPTS CHAPTER 16

SOLUTIONS TO CONCEPTS CHAPTER 16 . air = 30 m/. = 500 m/. Here S = 7 m So, t = t t = 330 500 SOLUIONS O CONCEPS CHPER 6 =.75 0 3 ec =.75 m.. Here gien S = 80 m = 60 m. = 30 m/ So the maximum time interal will be t = 5/ = 60/30 = 0.5 econd.

More information

Control of Wireless Networks with Flow Level Dynamics under Constant Time Scheduling

Control of Wireless Networks with Flow Level Dynamics under Constant Time Scheduling Control of Wirele Network with Flow Level Dynamic under Contant Time Scheduling Long Le and Ravi R. Mazumdar Department of Electrical and Computer Engineering Univerity of Waterloo,Waterloo, ON, Canada

More information

Turbulent Mixing and Chemical Reaction in Stirred Tanks

Turbulent Mixing and Chemical Reaction in Stirred Tanks Turbulent Mixing and Chemical Reaction in Stirred Tank André Bakker Julian B. Faano Blend time and chemical product ditribution in turbulent agitated veel can be predicted with the aid of Computational

More information

B) 40.8 m C) 19.6 m D) None of the other choices is correct. Answer: B

B) 40.8 m C) 19.6 m D) None of the other choices is correct. Answer: B Practice Test 1 1) Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.60 s and reaches the level of the top of the pole after a total elapsed time

More information

Support Vector Machine Based Electricity Price Forecasting For Electricity Markets utilising Projected Assessment of System Adequacy Data.

Support Vector Machine Based Electricity Price Forecasting For Electricity Markets utilising Projected Assessment of System Adequacy Data. The Sixth International Power Engineering Conference (IPEC23, 27-29 November 23, Singapore Support Vector Machine Baed Electricity Price Forecating For Electricity Maret utiliing Projected Aement of Sytem

More information

Lecture Presentation Chapter 2 Motion in One Dimension

Lecture Presentation Chapter 2 Motion in One Dimension Lecture Presentation Chapter 2 Motion in One Dimension Suggested Videos for Chapter 2 Prelecture Videos Motion Along a Line Acceleration Free Fall Video Tutor Solutions Motion in One Dimension Class Videos

More information

Bob York. Simple FET DC Bias Circuits

Bob York. Simple FET DC Bias Circuits Bob York Simple FET DC Bia Circuit Loa-Line an Q-point Conier the effect of a rain reitor in the comnon-ource configuration: Smaller + g D out KL: Thi i the equation of a line that can be uperimpoe on

More information