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


 Roland Arnold
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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 24] 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 218]
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 220] 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 228]
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 231] (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 233] 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 225] 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 231] (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 233] (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 239] 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 241] 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 273] 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 261] (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 idebyide 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 246] 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 247] (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 249] (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 251]
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 284] (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 254] 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 268] 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 278] 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
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