Solution: The equations that govern the motion are:

Transcription

1 Problem 13.1 In Eample 13., suppose that the vehicle is dropped from a height h = 6m. What is the downward velocit 1 s after it is released? What is its downward velocit just before it reaches the ground? The equations that govern the motion are: a = g = 9.81 m/s v = gt s = 1 gt + h h v = gt = (9.81 m/s )(1 s) = 9.81 m/s. The downward velocit is 9.81 m/s. We need to first determine the time at which the vehicle hits the ground s = = 1 h gt + h t = g = (6 m) 9.81 m/s = 1.16 s Now we can solve for the velocit v = gt = (9.81 m/s )(1.16 s) = 1.8 m/s. The downward velocit is 1.8 m/s. Problem 13. The milling machine is programmed so that during the interval of time from t = tot = s, the position of its head (in inches) is given as a function of time b s = 4t t 3. What are the velocit (in in/s) and acceleration (in in/s ) of the head at t = 1s? The motion is governed b the equations s = (4 in/s)t ( in/s )t, s v = (4 in/s) ( in/s )t, a = ( in/s ). At t = 1 s, we have v =,a = 4 in/s. 9

2 Problem 13.3 In an eperiment to estimate the acceleration due to gravit, a student drops a ball at a distance of 1 m above the floor. His lab partner measures the time it takes to fall and obtains an estimate of.46 s. What do the estimate the acceleration due to gravit to be? Let s be the ball s position relative to the floor. Using the value of the acceleration due to gravit that the obtained, and assuming that the ball is released at t =, determine s (in m) as a function of time. s The governing equations are a = g v = gt s = 1 gt + h s When the ball hits the floor we have = 1 gt + h g = h (1 m) = = 9.45 m/s t (.46 s) g = 9.45 m/s The distance s is then given b s = 1 (9.45 m/s ) + 1m. s = (4.73 m/s )t + 1. m. Problem 13.4 The boat s position during the interval of time from t = stot = 1 s is given b s = 4t + 1.6t.8t 3 m. Determine the boat s velocit and acceleration at t = 4s. What is the boat s maimum velocit during this interval of time, and when does it occur? s = 4t + 1.6t.8t 3 v = ds a = dv = t.4t = 3..48t a) v(4s) = 1.96 m/s a(4s) = 1.8 m/s b) a = 3..48t = t = 6.67s v(6.67s) = m/s 1

3 Problem 13.5 The rocket starts from rest at t = and travels straight up. Its height above the ground as a function of time can be approimated b s = bt + ct 3, where b and c are constants. At t = 1 s, the rocket s velocit and acceleration are v = 9 m/s and a = 8. m/s. Determine the time at which the rocket reaches supersonic speed (35 m/s). What is its altitude when that occurs? The governing equations are s = bt + ct 3, s v = bt + 3ct, a = b + 6ct. Using the information that we have allows us to solve for the constants b and c. (9 m/s) = b(1 s) + 3c(1 s), (8. m/s ) = b + 6c(1 s). Solving these two equations, we find b = 8.8 m/s, c =.177 m/s3. When the rocket hits supersonic speed we have (35 m/s) = (8.8 m/s )t + 3(.177 m/s3 )t t = 13. s. The altitude at this time is s = (8.8 m/s )(13. s) + (.177 m/s3 )(13. s)3 s = 194 m. Problem 13.6 The position of a point during the interval of time from t = to t = 6 s is given b s = 1 t 3 + 6t + 4t m. What is the maimum velocit during this interval of time, and at what time does it occur? What is the acceleration when the velocit is a maimum? s = 1 t 3 + 6t + 4t m dv = (it could be a minimum) da = 3 so we have a maimum. This occurs at t = 4 s. At this point v = 3 t + 1t + 4 m/s Maimum velocit occurs where a = Ma velocit is at t = 4 s. where v = 8 m/s and a = m/s a = 3t + 1 m/s Please purchase PDF Split-Merge on to remove this watermark. 11

4 Problem 13.7 The position of a point during the interval of time from t = tot = 3 seconds is s = 1 + 5t t 3 m. What is the maimum velocit during this interval of time, and at what time does it occur? What is the acceleration when the velocit is a maimum? The velocit is ds = 1t 3t. The maimum occurs when dv = 1 6t =, from which This is indeed a maimum, since d v = 6 <. The maimum velocit is v = [ 1t 3t ] = 8.33 m/s t=1.667 t = 1 6 = seconds. The acceleration is dv = when the velocit is a maimum. Problem 13.8 The rotating crank causes the position of point P as a function of time to be s =.4 sin (πt) m. P (c) Determine the velocit and acceleration of P at t =.375 s. What is the maimum magnitude of the velocit of P? When the magnitude of the velocit of P is a maimum, what is the acceleration of P? s s =.4 sin(πt) v = ds a = dv =.8π cos(πt) = 1.6π sin(πt) a) v(.375s) = m/s a(.375) = 11. m/s b) v ma =.8π =.513 m/s c) v ma t =,nπ a = Problem 13.9 For the mechanism in Problem 13.8, draw graphs of the position s, velocit v, and acceleration a of point P as functions of time for t s. Using our graphs, confirm that the slope of the graph of s is zero at times for which v is zero, and the slope of the graph of v is zero at times for which a is zero. 1

5 Problem 13.1 A seismograph measures the horizontal motion of the ground during an earthquake. An engineer analzing the data determines that for a 1-s interval of time beginning at t =, the position is approimated b s = 1 cos(πt) mm. What are the maimum velocit and maimum acceleration of the ground during the 1-s interval? The velocit is ds = (π)1 sin(πt) mm/s =.π sin(πt) m/s. The velocit maima occur at dv =.4π cos(πt) =, from which πt = (n 1)π, or t = n = 1,, 3,...M, where (n 1), 4 (M 1) 4 1 seconds. These velocit maima have the absolute value ds = [.π] =.68 m/s. (n 1) t= 4 The acceleration is d s =.4π cos(πt). The acceleration maima occur at d 3 s 3 = d v =.8π 3 sin(πt) =, from which πt = nπ, ort = n,n=, 1,,...K, where K 1 seconds. These acceleration maima have the absolute value dv nπ =.4π = 3.95 m/s. t= Problem In an assembl operation, the robot s arm moves along a straight horizontal line. During an interval of time from t = tot = 1 s, the position of the arm is given b s = 3t t 3 mm. Determine the maimum velocit during this interval of time. What are the position and acceleration when the velocit is a maimum? s s = 3t t 3 mm v = 6t 6t mm/s a = 6 1t mm/s da = 1 mm/s 3 Maimum velocit occurs when dv = a =. This occurs at = 6 1t or t = 1/ second. (since da/ <, we have a maimum). The velocit at this time is v = (6) v = 15 mm/s ( ) ( ) mm/s 4 The position and acceleration at this time are s = mm s = 5mm a = mm/s 13

6 Problem 13.1 In Active Eample 13.1, the acceleration (in m/s ) of point P relative to point O is given as a function of time b a = 3t. Suppose that at t = the position and velocit of P are s = 5 m and v = m/s. Determine the position and velocit of P at t = 4s. The governing equations are a = (3 m/s 4 )t v = 1 3 (3 m/s4 )t 3 + ( m/s) O s P s s = 1 1 (3 m/s4 )t 4 + ( m/s)t + (5 m) At t = 4 s, we have s = 77 m,v = 66 m/s. Problem The Porsche starts from rest at time t =. During the first 1 seconds of its motion, its velocit in km/h is given as a function of time b v =.8t.88t, where t is in seconds. What is the car s maimum acceleration in m/s, and when does it occur? What distance in km does the car travel during the 1 seconds?.8 km hr.88 km hr First convert the numbers into meters and seconds )( ) 1hr = 6.33 m/s 36 s ( 1 m 1km ( )( ) 1 m 1hr =.44 m/s 1km 36 s The governing equations are then s = 1 (6.33 m/s)(t /s) 1 3 (.88 m/s)(t3 /s ), v = (6.33 m/s)(t/s) (.88 m/s)(t /s ), a = (6.33 m/s ) (.88 m/s)(t/s ), d a = (.88 m/s)(1/s ) = 1.76 m/s 3. The maimum acceleration occurs at t = (and decreases linearl from its initial value). a ma = 6.33 t = In the first 1 seconds the car travels a distance [ ( 1 s =.8 km ) (1 s) 1 (.88 km ) (1 s) 3 ]( ) 1hr hr s 3 hr s 36 s s =.35 km. 14

7 Problem The acceleration of a point is a = t m/s. When t =,s = 4 m and v = 1 m/s. What are the position and velocit at t = 3s? v = a+ C 1, The velocit is where C 1 is the constant of integration. Thus v = t + C 1 = 1t + C 1. At t =, v = 1 m/s, hence C 1 = 1 and the velocit is v = 1t 1 m/s. The position is s = v+ C, At t =, s = 4 m, thus C = 4. The position is s = ( ) 1 t 3 1t + 4 m. 3 At t = 3 seconds, [ ] 1 s = 3 t3 1t + 4 = 1 m. t=3 The velocit at t = 3 seconds is v = [ 1t 1 ] = 8 m/s. t=3 where C is the constant of integration. s = (1t 1) + C = ( ) 1 t 3 1t + C. 3 Problem The acceleration of a point is a = 6t 36t m/s. When t =,s = and v = m/s. What are position and velocit as a function of time? The velocit is v = a+ C 1 = (6t 36t ) + C 1 = 3t 1t 3 + C 1. The position is s = v+ C = (3t 1t 3 + ) + C At t =, v = m/s, hence C 1 =, and the velocit as a function of time is v = 3t 1t 3 + m/s. = 1t 3 3t 4 + t + C. At t =, s =, hence C =, and the position is s = 1t 3 3t 4 + t m 15

8 Problem As a first approimation, a bioengineer studing the mechanics of bird flight assumes that the snow petrel takes off with constant acceleration. Video measurements indicate that a bird requires a distance of 4.3 m to take off and is moving at 6.1 m/s when it does. What is its acceleration? The governing equations are a = constant, v = at, s = 1 at. Using the information given, we have 6.1 m/s = at, 4.3 m= 1 at. Solving these two equations, we find t = 1.41 s and a = 4.33 m/s. Problem Progressivel developing a more realistic model, the bioengineer net models the acceleration of the snow petrel b an equation of the form a = C(1 + sin ωt), where C and ω are constants. From video measurements of a bird taking off, he estimates that ω = 18/s and determines that the bird requires 1.4 s to take off and is moving at 6.1 m/s when it does. What is the constant C? acceleration We find an epression for the velocit b integrating the a = C(1 + sin ωt), v = Ct + C ω (1 cos ωt) = C ( t + 1 ω 1 ω cos ωt ). Using the information given, we have 6.1 m/s = C (1.4 s + s 18 s ) 18 cos[18(1.4)] Solving this equation, we find C = 4.8 m/s. 16

9 Problem Missiles designed for defense against ballistic missiles have attained accelerations in ecess of 1 g s, or 1 times the acceleration due to gravit. Suppose that the missile shown lifts off from the ground and has a constant acceleration of 1 g s. How long does it take to reach an altitude of 3 m? How fast is it going when it ranches that altitude? The governing equations are a = constant, v = at, s = 1 at Using the given information we have 3 m = 1 1(9.81 m/s )t t =.47 s The velocit at that time is v = 1(9.81 m/s )(.47 s) = 43 m/s. v = 43 m/s. Problem Suppose that the missile shown lifts off from the ground and, because it becomes lighter as its fuel is epended, its acceleration (in g s) is given as a function of time in seconds b a = 1 1.t. What is the missile s velocit in kilometres per hour 1s after liftoff? We find an epression for the velocit b integrating the acceleration (valid onl for <t<5s). t t ( ) 1g 1gs v = a = = 1.(t/s). ln 1 1.[t/s] At time t = 1 s, we have v = 1(. ( 9 81 m/s )(1 s) ln. v = 394 k m/ h ) = 195 m/s. 17

10 Problem 13. The airplane releases its drag parachute at time t =. Its velocit is given as a function of time b 8 v = 1 +.3t m/s. What is the airplane s acceleration at t = 3s? v = t ; a = dv 5.6 = (1 +.3t) a(3 s) = 6.66 m/s Problem 13.1 How far does the airplane in Problem 13. travel during the interval of time from t = to t = 1 s? v = 8 1 s 1 +.3t ; s = 8 = 5 ln 1 +.3t ( ) = 359 m Problem 13. The velocit of a bobsled is v = 1t m/s. When t = s, the position is s = 5 m. What is its position at t = 1 s? The equation for straight line displacement under constant acceleration is s = a(t t ) + v(t )(t t ) + s(t ). Choose t =. At t =, the acceleration is [ ] dv(t) a = = 1 m/s, t= the velocit is v(t ) = 1() = m/s, and the initial displacement is s(t ) = 5 m. At t = 1 seconds, the displacement is s = 1 (1 ) + (1 ) + 5 = 55 m 18

11 Problem 13.3 In September, 3, Ton Schumacher started from rest and drove 4 km in seconds in a National Hot Rod Association race. His speed as he crossed the finish line was 58 km/h. Assume that the car s acceleration can be epressed b a linear function of time a = b + ct. Determine the constants b and c. What was the car s speed s after the start of the race? a = b + ct, v = bt + ct, s = bt + ct3 6 Both constants of integration are zero. 58 km/h = b(4.498 s) + c (4.498 s) 4 km = b (4.498 s) + c (4.498 s)3 6 b = 54 m/s 3 c = 9.5 m/s v = b( s) + c (s) = 89 m/s Problem 13.4 The velocit of an object is v = t m/s. When t = 3 seconds, its position is s = 6 m. What are the position and acceleration of the object at t = 6s? dv(t) = 4t m/s. The acceleration is At t = 6 seconds, the acceleration is a = 4 m/s. Choose the initial conditions at t = 3 seconds. The position is obtained from the velocit: 6 [ s(t t ) = v(t) + s(t ) = t ] t3 + 6 = 17 m. 3 Problem 13.5 An inertial navigation sstem measures the acceleration of a vehicle from t = tot = 6s and determines it to be a = +.1t m/s. At t =, the vehicle s position and velocit are s = 4 m, v = 4 m/s, respectivel. What are the vehicle s position and velocit at t = 6s? a = +.1t m/s v = 4 m/s s = 4 m Integrating v = v + t +.1t / s = v t + t +.1t 3 /6 + s Substituting the known values at t = 6s,weget v = 55.8 m/s s = m 19

12 Problem 13.6 In Eample 13.3, suppose that the cheetah s acceleration is constant and it reaches its top speed of 1 km/h in 5 s. What distance can it cover in 1 s? The governing equations while accelerating are a = constant, v = at, s = 1 at. Using the information supplied, we have ( ) 1 1 m = a(5 s) a = m/s 36 s The distance that he travels in the first 1 s (5 seconds accelerating and then the last 5 seconds traveling at top speed) is s = 1 ( ) ( ) (5 s) 1 1 m 6.67 m/s + (1 s 5s) = 5 m. 36 s s = 5 m. Problem 13.7 The graph shows the airplane s acceleration during its takeoff. What is the airplane s velocit when it rotates (lifts off) at t = 3 s? a 9 m/ s Velocit = Area under the curve v = 1 (3 m/s + 9 m/s )(5 s) + (9 m/s )(5 s) = 55 m/s 3 m/s 5 s 3 s t Problem 13.8 Determine the distance traveled during its takeoff b the airplane in Problem for t 5s ( ) ( ) a = 6 m/s t + (3 m/s 6 m/s ), v = 5s 5s t + (3 m/s )t ( ) 6 m/s t 3 s = 5s 6 + (3 m/s ) t v(5 s) = 3 m/s, s(5 s) = 6.5 m for 5 s t 3 s a = 9 m/s, v = (9 m/s )(t 5s) + 3 m/s, s = (9 m/s ) (t 5s) + (3 m/s)(t 5s) m s(3 s) = 365 m

13 Problem 13.9 The car is traveling at 48 km/h when the traffic light 9 m ahead turns ellow. The driver takes one second to react before he applies the brakes. 48 km/h After he applies the brakes, what constant rate of deceleration will cause the car to come to a stop just as it reaches the light? How long does it take the car to travel the 9 m? 9 m for t 1s a =, v= 48 km/h = m/s,s= ( m/s)t s(1 s) = m for t>1s a = c(constant), v = ct m/s, s = c t + ( m/s )t m At the stop we have 9 m = c t + ( m/s )t m = ct m/s a) c = m/s b) t = s Problem 13.3 The car is traveling at 48 km/h when the traffic light 9 m ahead turns ellow. The driver takes 1 s to react before he applies the accelerator. If the car has a constant acceleration of m/s and the light remains ellow for 5 s, will the car reach the light before it turns red? How fast is the car moving when it reaches the light? First, convert the initial speed into m/s. 48 km/h = m/s. At the end of the 5 s, the car will have traveled a distance [ ] 1 d = ( m/s )(1 s) + ( m/s )(5 s 1s) + ( m/s )(5 s 1s) = 8.65 m. When the light turns red, the driver will still be 7.35 m from the light. No. To find the time at which the car does reach the light, we solve [ ] 1 9 m = ( m/s)(1 s) + ( m/s )(t 1s) + ( m/s )(t 1s) t = s. The speed at this time is v = m/s + ( m/s ) (5. 34 s 1s) =.1 m/s. v = 79. k m/ h. 1

14 Problem A high-speed rail transportation sstem has a top speed of 1 m/s. For the comfort of the passengers, the magnitude of the acceleration and deceleration is limited to m/s. Determine the time required for a trip of 1 km. Strateg: A graphical approach can help ou solve this problem. Recall that the change in the position from an initial time t to a time t is equal to the area defined b the graph of the velocit as a function of time from t to t. Divide the time of travel into three intervals: The time required to reach a top speed of 1 m/s, the time traveling at top speed, and the time required to decelerate from top speed to zero. From smmetr, the first and last time intervals are equal, and the distances traveled during these intervals are equal. The initial time is obtained from v(t 1 ) = at 1, from which t 1 = 1/ = 5 s. The distance traveled during this time is s(t 1 ) = at1 / from which s(t 1) = (5) / = 5 m. The third time interval is given b v(t 3 ) = at =, from which t 3 = 1/ = 5 s. Check. The distance traveled is s(t 3 ) = a t 3 + 1t 3, from which s(t 3 ) = 5 m. Check. The distance traveled at top speed is s(t ) = = 95 m = 95 km. The time of travel is obtained from the distance traveled at zero acceleration: s(t ) = 95 = 1t, from which t = 95. The total time of travel is t total = t 1 + t + t 3 = = 15 s = 17.5 minutes. A plot of velocit versus time can be made and the area under the curve will be the distance traveled. The length of the constant speed section of the trip can be adjusted to force the length of the trip to be the required 1 km. Problem 13.3 The nearest star, Proima Centauri, is 4. light ears from the Earth. Ignoring relative motion between the solar sstem and Proima Centauri, suppose that a spacecraft accelerates from the vicinit of the Earth at.1 g (.1 times the acceleration due to gravit at sea level) until it reaches one-tenth the speed of light, coasts until it is time to decelerate, then decelerates at.1 g until it comes to rest in the vicinit of Proima Centauri. How long does the trip take? (Light travels at m/s.) The distance to Proima Centauri is d = (4. light - ear)(3 1 8 m/s)(365.4 da) = m. ( ) 864 s 1 da Divide the time of flight into the three intervals. The time required to reach.1 times the speed of light is t 1 = v a = 3 17 m/s.981 m/s = seconds. The distance traveled is s(t 1 ) = a t 1 + v()t + s(), where v() = and s() = (from the conditions in the problem), from which s(t 1 ) = m. From smmetr, t 3 = t 1, and s(t 1 ) = s(t 3 ). The length of the middle interval is s(t ) = d s(t 1 ) s(t 3 ) = m. The time of flight at constant velocit is t = m = seconds. The total time of flight is t total = t 1 + t + t 3 = seconds. In solar ears: t total = ( sec ) ( )( ) 1 solar ears 1 das das 864 sec = 51.9 solar ears

15 Problem A race car starts from rest and accelerates at a = 5 + t m/s for 1 seconds. The brakes are then applied, and the car has a constant acceleration a = 3 m/s until it comes to rest. Determine the maimum velocit, the total distance traveled; (c) the total time of travel. (c) the total time of travel is t = 15. The total distance traveled is For the first interval, the velocit is v(t) = (5 + t) + v() = 5t + t since v() =. The velocit is an increasing monotone function; hence the maimum occurs at the end of the interval, t = 1 s, from which s(t 1) = a (t 1) + v(1)(t 1) + s(1), from which s(5) = (5) = m v ma = 15 m/s. The distance traveled in the first interval is s(1) = [ 1 5 (5t + t ) = t + 1 ] 1 3 t3 The time of travel in the second interval is = m. v(t 1) = = a(t 1) + v(1), t 1 s, from which (t 1) = 15 3 = 5, and Problem When t =, the position of a point is s = 6 m and its velocit is v = m/s. From t = tot = 6 s, the acceleration of the point is a = + t m/s. From t = 6 s until it comes to rest, its acceleration is a = 4 m/s. What is the total time of travel? What total distance does the point move? For the first interval the velocit is v(t) = ( + t ) + v() = [t + 3 ] t3 + m/s. The total time of travel is t total = = 45.5 seconds. The velocit at the end of the interval is v(6) = 158 m/s. The displacement in the first interval is ( s(t) = t + ) [ 3 t = t + 1 ] 6 t4 + t + 6. The distance traveled is s(t 6) = 4 (t 6) + v(6)(t 6) + s(6) = (39.5) + 158(39.5) + 7, The displacement at the end of the interval is s(6) = 7 m. For the second interval, the velocit is v(t 6) = a(t 6) + v(6) =,t 6, from which (t 6) = v(6) a = = from which the total distance is s total = 339 m 3

16 Problem Zoologists studing the ecolog of the Serengeti Plain estimate that the average adult cheetah can run 1 km/h and that the average springbuck can run 65 km/h. If the animals run along the same straight line, start at the same time, and are each assumed to have constant acceleration and reach top speed in 4 s, how close must the a cheetah be when the chase begins to catch a springbuck in 15 s? The top speeds are V c = 1 km/h = 7.78 m/s for the cheetah, and V s = 65 km/h = 18.6 m/s. The acceleration is a c = V c 4 = 6.94 m/s for the cheetah, and a s = V s 4 = m/s for the springbuck. Divide the intervals into the acceleration phase and the chase phase. For the cheetah, the distance traveled in the first is s c (t) = 6.94 (4) = m. The total distance traveled at the end of the second phase is s total = V c (11) = m. For the springbuck, the distance traveled during the acceleration phase is s s (t) = (4) = m. The distance traveled at the end of the second phase is s s (t) = 18.6(11) = 34.7 m. The permissible separation between the two at the beginning for a successful chase is d = s c (15) s s (15) = = 16.4 m Problem Suppose that a person unwisel drives 1 km/h in a 88 km/h zone and passes a police car going 88 km/h in the same direction. If the police officers begin constant acceleration at the instant the are passed and increase their speed to 19 km/h in 4 s, how long does it take them to be even with the pursued car? The conversion from mi/h to m/s is The distance traveled b the pursued car during this acceleration is km h = 1 m 36 s =.78 m/s The acceleration of the police car is a = ( )(. 78 ) m/s 4s =.85 m/s. The distance traveled during acceleration is s(t 1 ) =. 85 (4) + 88(. 78)(4) = 11 m. s c (t 1 ) = 1 (. 78 ) t 1 = (4) = m. The separation between the two cars at 4 seconds is d = = 1.4 m. This distance is traversed in the time 1.4 t = = ( )(. 78 ) The total time is t total = = 5.9 seconds. Problem If θ = 1 rad and dθ is the velocit of P relative to O? = 1 rad/s, what Strateg: You can write the position of P relative to O as s = ( m) cos θ + ( m) cos θ and then take the derivative of this epression with respect to time to determine the velocit. The distance s from point O is s = ( m) cos θ + ( m) cos θ. O m s m P The derivative is ds = 4 sin θ dθ. For θ = 1 radian and dθ = 1 radian/second, ds = v(t) = 4(sin(1 rad)) = 4(.841) = 3.37 m/s 4

17 Problem In Problem 13.37, if θ = 1 rad, dθ/ = rad/s and d θ/ =, what are the velocit and acceleration of P relative to O? The velocit is ds = 4 sin θ dθ = 4(sin(1 rad))( ) = 6.73 m/s. The acceleration is d s = 4 cos θ ( ) dθ 4 sin θ ( d ) θ, from which d s = a = 4 cos(1 rad)(4) = 8.64 m/s Problem If θ = 1 rad and dθ is the velocit of P relative to O? = 1 rad/s, what mm 4 mm The acute angle formed b the 4 mm arm with the horizontal is given b the sine law: sin α = 4 sin θ, O s P from which sin α = ( ) sin θ. 4 For θ = 1 radian, α =.4343 radians. The position relative to O is. s = cos θ + 4 cos α. The velocit is ds = v(t) = sin θ ( ) ( ) dθ dα 4 sin α. From the epression for the angle α, cos α ( ) dα =.5 cos θ ( ) dθ, from which the velocit is v(t) = ( sin θ tan α cos θ) ( ) dθ. Substitute: v(t) = 18.4 mm/s. 5

18 Problem 13.4 In Active Eample 13.4, determine the time required for the plane s velocit to decrease from 5 m/s to 1 m/s. From Active Eample 13.4 we know that the acceleration is given b a = (.4/m)v. We can find an epression for the velocit as a function of time b integrating a = dv 1 m/s 5 m/s = (.4/m)v dv v = (.4/m)t ( dv v = 1 ) 1 m/s ( = 1s v 5 m/s 1 m + 1s ) ( = 4s ) = (.4/m)t 5 m 5 m ( t = 4s )( ) 1m = s. 5 m.4 t = s. Problem An engineer designing a sstem to control a router for a machining process models the sstem so that the router s acceleration (in cm/s ) during an interval of time is given b a =.4v, where v is the velocit of the router in cm/s. When t =, the position is s = and the velocit is v = cm/s. What is the position at t = 3s? We will first find the velocit at t = 3. a = dv ( ).4 = v, s s v dv 3 s ( ).4 cm/s v =, s ( ) v ln = cm/s (.4 s ) (3 s) = 1. v = ( cm/s)e 1. =.6 cm/s. Now we can find the position a = vdv ( ).4 = v, ds s.6 cm/s cm/s vdv v ( ).4 s = ds s ( ).4 (.6 cm/s) ( cm/s)= s s s = cm/s = 3.49 cm.4/ s s = 3.49 cm 6

19 Problem 13.4 The boat is moving at 1 m/s when its engine is shut down. Due to hdrodnamic drag, its subsequent acceleration is a =.5v m/s, where v is the velocit of the boat in m/s. What is the boat s velocit 4 s after the engine is shut down? a = dv v 1 m/s = (.5 m 1 )v dv t v = (.5 m 1 ) 1 m/s v = 1 + (.5 s 1 )t 1 v v = (.5 m 1 )t 1 m/s v(4 s) = 3.33 m/s Problem In Problem 13.4, what distance does the boat move in the 4 s following the shutdown of its engine? From Problem 13.4 we know v = ds 1 m/s 4s 1 m/s = 1 + (.5 s 1 s(4 s) = )t 1 + (.5 s 1 )t [ ] + (1 s 1 )(4 s) s(4 s) = ( m) ln = 1.97 m Problem A steel ball is released from rest in a container of oil. Its downward acceleration is a =.4.6v cm/s, where v is the ball s velocit in cm/s.what is the ball s downward velocit s after it is released? a = dv v = (.4 cm/s) (.6 s 1 )v dv (.4 cm /s) (.6 s 1 )v = 5 3 ln ( v + 4 cm/s 4 cm/s t ) = t v = (4 cm/s) (1 ) e (.6 s 1 )t v( s) =.795 cm/s Problem In Problem 13.44, what distance does the ball fall in the first s after its release? From we know v = ds = (4 cm/s) (1 ) e (.6 s 1 )t t ( ) s( s) = (4 cm/s) 1 e (.6 s 1 )t = cm 3 s( s) = 3.34 cm. ( ) e (.6 s 1 )t 1 + (4 cm/s )t 7

20 Problem The greatest ocean depth et discovered is the Marianas Trench in the western Pacific Ocean. A steel ball released at the surface requires 64 minutes to reach the bottom. The ball s downward acceleration is a =.9g cv, where g = 9.81 m/s and the constant c = 3. s 1. What is the depth of the Marianas Trench in kilometers? a = dv =.9g cv. Separating variables and integrating, v dv t.9g cv = = t. Integrating and solving for v, v = ds =.9g c (1 e ct ). Integrating, s t ds = We obtain s =.9g c.9g c (1 e ct ). (t + e ct 1 c c ). At t = (64)(6) = 384 s, we obtain s = 11,5 m. Problem The acceleration of a regional airliner during its takeoff run is a = 14.3v m/s, where v is its velocit in m/s. How long does it take the airliner to reach its takeoff speed of m/s? a = dv = (14 m/s ) (.3 m 1 )v m/s dv t (14 m/s ) (.3 m 1 )v = t = 5.1 s Problem In Problem 13.47, what distance does the airliner require to take off? a = v dv ds = (14 m/s ) (.3 m 1 )v m/s vdv (14 m/s ) (.3 m 1 )v = s ds s = 343 m 8

21 Problem A sk diver jumps from a helicopter and is falling straight down at 3 m/s when her parachute opens. From then on, her downward acceleration is approimatel a = g cv, where g = 9.81 m/s and c is a constant. After an initial transient period she descends at a nearl constant velocit of 5 m/s. (c) What is the value of c, and what are its SI units? What maimum deceleration is the sk diver subjected to? What is her downward velocit when she has fallen meters from the point at which her parachute opens? Assume c>. After the initial transient, she falls at a constant velocit, so that the acceleration is zero and cv = g, from which Integrate: ( 1 ) ln g cv =s + C. c (c) c = g 9.81 m/s = v (5) m =.394 m 1 /s The maimum acceleration (in absolute value) occurs when the parachute first opens, when the velocit is highest: a ma = g cv = g c(3) =343.4 m/s Choose coordinates such that distance is measured positive downward. The velocit is related to position b the chain rule: dv = dv ds = v dv ds ds = a, from which vdv g cv = ds. When the parachute opens s = and v = 3 m/s, from which ( ) 1 C = ln g 9c = c The velocit as a function of distance is ln g cv = c(s + C). For s = m, v = 14.4 m/s Problem 13.5 The rocket sled starts from rest and accelerates at a = 3 + t m/s until its velocit is 4 m/s. It then hits a water brake and its acceleration is a =.3v m/s until its velocit decreases to 1 m/s. What total distance does the sled travel? a = 3 + t m/s v = 3t + t m/s s = 15t + t 3 /3m Acceleration Phase When v = 4 m/s, acceleration ends. At this point, t = 1 s and s = 1833 m. Deceleration Phase starts at s 1 = 1833 m, v 1 = 4 m/s. Let us start a new clock for the deceleration phase. v f = 1 m/s a = v dv ds =.3v sf s 1 ds = 1 (.3) vf v 1 vdv v s f 1833 m = 1 [ln(1) ln(4)] (.3) s f = 3 m 9

22 Problem In Problem 13.5, what is the sled s total time of travel? From the solution to Problem 13.5, the acceleration takes 1 s. At t = 1 s, the velocit is 4 m/s. We need to find out how long it takes to decelerate from 4 m/s to 1 m/s and add this to the 1 s required for acceleration. The deceleration is given as a = dv td.3 = =.3v m/s 1 4 dv v.3t d = 1 1 v 4 t d =.5 s.3t d = 3 4 ( 1 = 1 1 ) 4 t = 1 + t d = 1.5 s Problem 13.5 A car s acceleration is related to its position b a =.1s m/s. When s = 1 m, the car is moving at 1 m/s. How fast is the car moving when s = 4 m? a = v dv =.1s m/s ds vf 1 4 vdv=.1 sds 1 [ v ] vf [ s ] 4 m =.1 1 m/s 1 m vf = (4 1 ) v f = 4.5 m/s Problem Engineers analzing the motion of a linkage determine that the velocit of an attachment point is given b v = A + 4s m/s, where A is a constant. When s = m, its acceleration is measured and determined to be a = 3 m/s. What is its velocit of the point when s = m? The velocit as a function of the distance is When s = m, a = 3 m/s, from which A = 4. v dv ds = a. Solve for a and carr out the differentiation. The velocit at s = mis v = 4 + 4( ) = m/s a = v dv ds = (A + 4s )(8s). 3

23 Problem The acceleration of an object is given as a function of its position in feet b a = s ( m/s ). When s =, its velocit is v = 1 m/s. What is the velocit of the object when s = m? We are given a = vdv ( = ds ) s, m-s v 1m/s ( m vdv = ) s ds m-s v (1 m/s) v = 3.4 m/s. ( (m) 3 = m-s) 3 Problem Gas guns are used to investigate the properties of materials subjected to high-velocit impacts. A projectile is accelerated through the barrel of the gun b gas at high pressure. Assume that the acceleration of the projectile is given b a = c/s, where s is the position of the projectile in the barrel in meters and c is a constant that depends on the initial gas pressure behind the projectile. The projectile starts from rest at s = 1.5 m and accelerates until it reaches the end of the barrel at s = 3 m. Determine the value of the constant c necessar for the projectile to leave the barrel with a velocit of m/s. s a = v dv ds = c s, ( m/s) m/s ( ) 3m = c ln 1.5 m 3m c vdv = 1.5m s ds c = m /s Problem If the propelling gas in the gas gun described in Problem is air, a more accurate modeling of the acceleration of the projectile is obtained b assuming that the acceleration of the projectile is given b a = c/s γ, where γ = 1.4 is the ratio of specific heats for air. (This means that an isentropic epansion process is assumed instead of the isothermal process assumed in Problem ) Determine the value of the constant c necessar for the projectile to leave the barrel with a velocit of m/s. a = v dv ds = c s 1.4, ( m/s) m/s 3m c vdv = ds 1.5m s1.4 =.5c ((3m).4 (1.5m).4) c = m.4 /s 31

24 Problem A spring-mass oscillator consists of a mass and a spring connected as shown. The coordinate s measures the displacement of the mass relative to its position when the spring is unstretched. If the spring is linear, the mass is subjected to a deceleration proportional to s. Suppose that a = 4s m/s, and that ou give the mass a velocit v = 1 m/s in the position s =. s How far will the mass move to the right before the spring brings it to a stop? What will be the velocit of the mass when it has returned to the position s =? The velocit of the mass as a function of its position is given b vdv/ds = a. Substitute the given acceleration, separate variables, and integrate: vdv= 4sds, from which v / = s + C. The initial velocit v() = 1 m/s at s =, from which C = 1/. The velocit is v / = s + 1/. The velocit is zero at the position given b = (s 1 ) + 1, 1 from which s 1 =± 4 =±1 m. Since the displacement has the same sign as the velocit, s 1 =+1/ m is the distance traveled before the spring brings it to a stop. At the return to s =, the velocit is v =± =±1 m/s. From the phsical situation, the velocit on the first return is negative (opposite the sign of the initial displacement), v = 1 m/s. Problem In Problem 13.57, suppose that at t = ou release the mass from rest in the position s = 1 m. Determine the velocit of the mass as a function of s as it moves from the initial position to s =. From the solution to Problem 13.57, the velocit as a function of position is given b v = s + C. At t =,v = and s = 1 m, from which C = (1) =. The velocit is given b v(s) =±( 4s + 4) 1 =± 1 s m/s. From the phsical situation, the velocit is negative (opposite the sign of the initial displacement): v = 1 s m/s. [Note: From the initial conditions, s 1 alwas.] 3

25 Problem A spring-mass oscillator consists of a mass and a spring connected as shown. The coordinate s measures the displacement of the mass relative to its position when the spring is unstretched. Suppose that the nonlinear spring subjects the mass to an acceleration a = 4s s 3 m/s and that ou give the mass a velocit v = 1 m/s in the position s =. How far will the mass move to the right before the springs brings it to a stop? What will be the velocit of the mass when it has returned to the position s =? s Find the distance when the velocit is zero. a = vdv ( ) ( ) 4 = ds s s + m s s 3 d vdv = 1m/s (1 m/s) = [( ) ( ) ] 4 s s + m s s 3 ds ( ) 4 d ( ) d 4 s m s 4 Solving for d we find d =.486 m. When the cart returns to the position s =, we have a = vdv ( ) ( ) 4 = ds s s + m s s 3 v [( ) ( ) ] 4 vdv = 1m/s s s + m s s 3 ds = v (1 m/s) =, v =±1 m/s. The cart will be moving to the left, so we choose v = 1 m/s. Problem 13.6 The mass is released from rest with the springs unstretched. Its downward acceleration is a = 3. 5s m/s, where s is the position of the mass measured from the position in which it is released. How far does the mass fall? What is the maimum velocit of the mass as it falls? The acceleration is given b s a = dv = dv ds = v dv ds ds = 3. 5s m/s. Integrating, we get v s vdv= (3. 5s)ds or v = 3.s 5s. The mass falls until v =. Setting v =, we get = (3. 5s)s. Wefind v = ats = and at s = 1.88 m. Thus, the mass falls 1.88 m before coming to rest. From the integration of the equation of motion, we have v = (3.s 5s ). The maimum velocit occurs where dv ds =. From the original equation for acceleration, we have a = v dv ds = (3. 5s) m/s. Since we want maimum velocit, we can assume that v = at this point. Thus, = (3. 5s), ors = (3./5) m when v = v MAX. Substituting this value for s into the equation for v, weget ( (3.) vmax = 5 or v MAX = 4.55 m/s ) (5)(3.) 5, 33

26 Problem Suppose that the mass in Problem 13.6 is in the position s = and is given a downward velocit of 1 m/s. How far does the mass fall? What is the maimum velocit of the mass as it falls? a = v dv ds = (3. m/s ) (5 s )s v s vdv = [(3. m/s ) (5 s )s]ds 1 m/s v (1 m/s) = (3. m/s ) s (5 s ) s v = (1 m/s) + (64.4 m/s )s (5 s )s The mass falls until v = = (1 m/s) + (64.4 m/s )s (5 s )s s =. m The maimum velocit occurs when a = = (3. m/s ) (5 s )s s =.644 m v = (1 m/s) + (64.4 m/s )(.644 m) (5 s )(.644 m) v = 1.99 m/s Problem 13.6 If a spacecraft is 161 km above the surface of the earth, what initial velocit v straight awa from the earth would be required for the vehicle to reach the moon s orbit 38, 94 km from the center of the earth? The radius of the earth is 637 km. Neglect the effect of the moon s gravit. (See Eample 13.5.) 161 km 38,94 km For computational convenience, convert the acceleration due to Earth s gravit into the units given in the problem, namel miles and hours: g = ( m 1 s )( )( 1 km 36 s ) 1h = km/h. 1 m The velocit as a function of position is given b v dv ds = a = gr E s. Separate variables, vdv= gre ds s. Integrate: v = gr E ( 1 ) + C. Suppose that the velocit at the distance of the Moon s orbit is zero. Then ( 637 ) = ( ) + C, 3894 from which C = km /h. At the 161 km altitude, the equation for the velocit is ( ) v = g RE + C. R E From which v = = 39,413 km/h Converting: ( )( )( ) km 1 m 1h v = = 1, 948 m/s. 1h 1km 36 s Check: Use the result of Eample 13.5 v = gr E (where H >s alwas), and H = 38, 94, ( 1 1 ), s H from which v = 39,413 km/h. check. 34

27 Problem The moon s radius is 1738 km. The magnitude of the acceleration due to gravit of the moon at a distance s from the center of the moon is m/s. s Suppose that a spacecraft is launched straight up from the moon s surface with a velocit of m/s. Set G = m 3 /s, r = m, v = m/s a = v dv v r ds = G s G v vdv = ds v s v ( 1 = G r 1 ) r ( ) v = v r r + G rr r What will the magnitude of its velocit be when it is 1 km above the surface of the moon? What maimum height above the moon s surface will it reach? v(r m) = 1395 m/s The maimum velocit occurs when v = r = G = 61 km G r v h = r r = 47 km Problem 13.64* The velocit of an object subjected onl to the earth s gravitational field is [ ( 1 v = v + gr E s 1 )] 1/, s where s is the object s position relative to the center of the earth, v is the velocit at position s, and R E is the earth s radius. Using this equation, show that the object s acceleration is given as a function of s b a = gre /s. ( 1 v = v + gr E s 1 ) 1/ s a = dv = v dv ds Rewrite the equation given as v = v + gr E s gr E s Take the derivative with respect to s. v dv ds = gr E s Thus a = v dv ds gr E s Problem Suppose that a tunnel could be drilled straight through the earth from the North Pole to the South Pole and the air was evacuated. An object dropped from the surface would fall with the acceleration a = gs/r E, where g is the acceleration of gravit at sea level, R E is radius of the earth, and s is the distance of the object from the center of the earth. (The acceleration due to gravitation is equal to zero at the center of the earth and increases linearl with the distance from the center.) What is the magnitude of the velocit of the dropped object when it reaches the center of the earth? s N S R E Tunnel v dv ds = gs R E. The velocit as a function of position is given b Separate variables and integrate: ( ) g v = s + C. R E At s = R E, v =, from which C = gr E. Combine and reduce: ( ) v = gr E 1 s RE At the center of the earth s =, and the velocit is v = gr E 35

28 Problem Determine the time in seconds required for the object in Problem to fall from the surface of the earth to the center. The earth s radius is 637 km. From Problem 13.65, the acceleration is a = v dv ds = g s R E v s ( ) g vdu= sds R E R E ( ) g v = (RE R s ) E Recall that v = ds/ v = ds g =± RE R s E R E ds g =± RE s R E tf ( ) g s t f =±sin 1 =±sin 1 (1) R E R E R E RE π t f =± =±166 s =±1.1 min g Problem In a second test, the coordinates of the position (in m) of the helicopter in Active Eample 13.6 are given as functions of time b = 4 + t, = 4 + 4t + t. What is the magnitude of the helicopter s velocit at t = 3s? What is the magnitude of the helicopter s acceleration at t = 3s? We have = (4 m) + ( m/s)t, = (4 m) + (4 m/s)t + (1 m/s )t, v = ( m/s), v = (4 m/s) + (1 m/s )t, a =, a = (1 m/s ). At t = 3 s, we have = 1 m, v = m/s, a =, = 5 m, v = 1 m/s, a = m/s. Thus v = v + v = 1. m/s v = 1. m/s. a = + ( m/s ) = m/s a = m/s. 36

29 Problem In terms of a particular reference frame, the position of the center of mass of the F-14 at the time shown (t = ) is r = 1i + 6j + k (m). The velocit from t = tot = 4sisv = (5 + 6t)i + (1 + t )j (4 + t )k (m/s). What is the position of the center of mass of the plane at t = 4s? r = 1i + 6j + k m v = (5 + 6t)i + (1 + t )j (4 + t )k m/s 4 4 = v = 5t + 3t + 4 = (5)(4) + 3(4) + 1 m = 66. m 4 4 = v = 1t + t 3 /3 + 4 = 1(4) + (4) 3 /3 + 6m= 75.3 m 4 z 4 = v z = (4t + t 3 /3) + z z 4 = 4(4) (4) 3 /3 + = 36.7 m r t=4s = 66i j 36.7k (m) Problem In Eample 13.7, suppose that the angle between the horizontal and the slope on which the skier lands is 3 instead of 45. Determine the distance d to the point where he lands. 3 m d 45 The skier leaves the surface at 1 m/s. The equations are a =, a = 9.81 m/s, v = (1 m/s) cos, v = (9.81 m/s )t (1 m/s) sin s = (1 m/s) cos t, s = 1 (9.81 m/s )t (1 m/s) sin t When he hits the slope, we have s = d cos 3 = (1 m/s) cos t, s = ( 3 m) d sin 3 = 1 (9.81 m/s )t (1 m/s) sin t Solving these two equations together, we find t = 1.1 s, d = 11. m. 37

30 Problem 13.7 A projectile is launched from ground level with initial velocit v = m/s. Determine its range R if θ = 3 ; θ = 45 (c) θ = 6. Set g = 9.81 m/s,v = m/s a = g, v = gt + v sin θ, s = 1 gt + v sin θ t a =, v = v cos θ, s = v cos θ t u R When it hits the ground, we have = 1 gt + v sin θ t t = v sin θ g R = v cos θ t R = v sin θ g a) θ = 3 R = 35.3 m b) θ = 45 R = 4.8 m c) θ = 6 R = 35.3 m Problem Immediatel after the bouncing golf ball leaves the floor, its components of velocit are v =.66 m/s and v = 3.66 m/s. Determine the horizontal distance from the point where the ball left the floor to the point where it hits the floor again. The ball leaves the floor at =, =. Determine the ball s coordinate as a function of. (The parabolic function ou obtain is shown superimposed on the photograph of the ball.) The governing equations are a =, a = g, v = v, v = gt + v, = v t, = 1 gt + v t When it hits the ground again = 1 gt + v t t = v g (.66 m/s)(3.66 m/s) = 9.81 m/s =.494 m. At an point of the flight we have t =, = 1 ( ) ( ) v g + v v v ( ) = m/s [.66 m/s] 3.66 m/s +.66 m/s ( ) = v t = v = v v g g ( ) 11. = m 38

31 Problem 13.7 Suppose that ou are designing a mortar to launch a rescue line from coast guard vessel to ships in distress. The light line is attached to a weight fired b the mortar. Neglect aerodnamic drag and the weight of the line for our preliminar analsis. If ou want the line to be able to reach a ship 91 m awa when the mortar is fired at 45 above the horizontal, what muzzle velocit is required? 45 From 13.7 we know that R = v sin θ Rg v = g sin θ v = ( 91)( 9.81 m/s ) sin(9 ) = 9.9 m/s Problem In Problem 13.7, what maimum height above the point where it was fired is reached b the weight? From Problem 13.7 we have v = gt + v sin θ, s = 1 gt + v sin θ t When we reach the maimum height, = gt + v sin θ t = v sin θ g h = 1 gt + v sin θ t h = 1 g ( v sin θ g ) ( ) v sin θ + v sin θ g h = v sin θ g Putting in the numbers we have h = ( 9.9 m/s ) sin (45 ) ( =.78 m 9.81 m/s ) 39

32 Problem When the athlete releases the shot, it is 1.8 m above the ground and its initial velocit is v = 13.6 m/s. Determine the horizontal distance the shot travels from the point of release to the point where it hits the ground. 3 v The governing equations are a =, v = v cos 3, s = v cos 3 t, a = g v = gt + v sin 3 s = 1 gt + v sin 3 t + h When it hits the ground, we have s = (13.6 m/s) cos 3 t, s = = 1 (9.81 m/s )t + (13.6 m/s) sin 3 t m. Solving these two equations, we find t = 1.6 s, s = 19. m. Problem A pilot wants to drop surve markers at remote locations in the Australian outback. If he flies at a constant velocit v = 4 m/s at altitude h = 3 m and the marker is released with zero velocit relative to the plane, at what horizontal d from the desired impact point should the marker be released? h We want to find the horizontal distance traveled b the marker before it strikes the ground ( goes to zero for t>.) a = a = g d v = v v = v gt = + v t = + v t gt / From the problem statement, =,v =,v = 4 m/s, and = 3 m The equation for becomes = 3 (9.81)t / Solving with =, we get t f =.47 s. Substituting this into the equation for, we get f = 4t f = 98.9 m 4 c 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copright laws as the currentl eist. No portion of this material ma be reproduced, in an form or b an means, without permission in writing from the publisher.

33 Problem If the pitching wedge the golfer is using gives the ball an initial angle θ = 5, what range of velocities v will cause the ball to land within 3 m of the hole? (Assume the hole lies in the plane of the ball s trajector). v θ 3 m 3 m Set the coordinate origin at the point where the golfer strikes the ball. The motion in the horizontal () direction is given b a =, V = V cos θ, = (V cos θ )t. The motion in the vertical () direction is given b a = g, V = V sin θ gt, = (V sin θ )t gt. From the equation, we can find the time at which the ball reaches the required value of (7 or 33 metres). This time is t f = f /(V cos θ ). We can substitute this information the equation for Y with Y f = 3m and solve for V. The results are: For hitting (7,3) metre, V = 31. m/s. For hitting (33,3) metre, V = 34. m/s. Problem A batter strikes a baseball.9 m above home plate and pops it up. The second baseman catches it 1.8 m above second base 3.68 s after it was hit. What was the ball s initial velocit, and what was the angle between the ball s initial velocit vector and the horizontal? Second base 7 m The equations of motion g = 9.81 m/s a = a = g v = v cos θ v = gt + v sin θ s = v cos θ t s = 1 gt + v sin θ t +.9m When the second baseman catches the ball we have 38. m = v cos θ (3.68 s) Home plate 7 m 1.8m= 1 ( 9.81 m/s ) (3.68 s) + v sin θ (3.68 s) Solving simultaneousl we find v = 1 m/s, θ =

34 Problem A baseball pitcher releases a fastball with an initial velocit v = km/h. Let θ be the initial angle of the ball s velocit vector above the horizontal. When it is released, the ball is 1.83 m above the ground and 17.7 m from the batter s plate. The batter s strike zone etends from.56 m above the ground to 1.37 m above the ground. Neglecting aerodnamic effects, determine whether the ball will hit the strike zone if θ = 1 ; if θ = m.56 m 17.7m The initial velocit is v = km/h = 4.3 m/s. The velocit equations are (1) () (3) dv dv =, from which v = v cos θ. = g, from which v = gt + v sin θ. d = v cos θ, from which (t) = v cos θt, since the initial position is zero. Substitute: (t p ) = h = g ( ) d + d tan θ v cos θ For θ = 1, h = 1. m,yes, the pitcher hits the strike zone. For θ =, h = 1. 5 m No, the pitcher misses the strike zone. (4) d = gt + v sin θ, from which (t) = g t + v sin θt , since the initial position is () = 1.83 m. At a distance d = 17.7 m, the height is h. The time of passage across the home plate is (t p ) = d = v cos θt p, from which t p = d v cos θ. Problem In Problem 13.78, assume that the pitcher releases the ball at an angle θ = 1 above the horizontal, and determine the range of velocities v (in m/s) within which he must release the ball to hit the strike zone. From the solution to Problem 13.78, h = g ( ) d + d tan θ , v cos θ where d = 17.7 m, and 1.37 h.56 m. Solve for the initial velocit: gd v = cos θ(d tan θ h). For h = 1. 37, v = m/s. For h =. 56 m, v = m/s. The pitcher will hit the strike zone for velocities of release of v m/s, and a release angle of θ = 1. Check: The range of velocities in miles per hour is 11. 1km/h v km/h, which is within the range of major league pitchers, although the 16.9 km upper value is achievable onl b a talented few (Nolan Ran, while with the Houston Astros, would occasionall in a game throw a km fast ball, as measured b hand held radar from behind the plate). 4

35 Problem 13.8 A zoolog student is provided with a bow and an arrow tipped with a sringe of sedative and is assigned to measure the temperature of a black rhinoceros (Diceros bicornis). The range of his bow when it is full drawn and aimed 45 above the horizontal is 1 m. A truculent rhino charges straight toward him at 3 km/h. If he full draws his bow and aims above the horizontal, how far awa should the rhino be when the student releases the arrow? The strateg is to determine the range and flight time of the arrow when aimed above the horizontal, to determine the distance traveled b the rhino during this flight time, and then (c) to add this distance to the range of the arrow. Neglect aerodnamic drag on the arrow. The equations for the trajector are: Denote the constants of integration b V,V,C,C, and the velocit of the arrow b V A. (1) dv =, from which v = V.Att =,V = V A cos θ. () (3) (4) dv = g, from which v = gt + V. At t =,V = V A sin θ. d = v = V A cos θ, from which (t) = V A cos θt + C. At t =, () =, from which C =. d = v = gt + V A sin θ, from which = g t + V A sin θt + C. At t =, =, from which C =. The time of flight is given b ( (t flight ) = = g ) t flight + V A sin θ t flight, from which t flight = V A sin θ. g The range is given b (t flight ) = R = V A cos θt flight = V A cos θ sin θ. g The maimum range (1 meters) occurs when the arrow is aimed 45 above the horizon. Solve for the arrow velocit: V A = grma = 31.3 m/s. The time of flight when the angle is is t flight = V A sin θ g =.18 s, and the range is R = V A cos θt flight = 64.3 m. The speed of the rhino is 3 km/h = 8.33 m/s. The rhino travels a distance d = 8.33(.18) = 18. m. The required range when the arrow is released is d + R = 8.5 m. 43

36 Problem The crossbar of the goalposts in American football is c = 3.5 m above the ground. To kick a field goal, the ball must make the ball go between the two uprights supporting the crossbar and be above the crossbar when it does so. Suppose that the kicker attempts a m field goal, and kicks the ball with an initial velocit v = 1.3 m/s and θ = 4. B what vertical distance does the ball clear the crossbar? v θ c c Set the coordinate origin at the point where the ball is kicked. The (horizontal) motion of the ball is given b a =, V = V cos θ, = (V cos θ )t. The motion is given b a = g, V = V sin θ gt, = (V sin θ )t gt. Set = c = m and find the time t c at which the ball crossed the plane of the goal posts. Substitute this time into the equation to find the coordinate Y B of the ball as it passes over the crossbar. Substituting in the numbers (g = 9.81 m/s ),wegett c =.4 s and B = m. Thus, the ball clears the crossbar b 3.7 m. Problem 13.8 An American football quarterback stands at A. At the instant the quarterback throws the football, the receiver is at B running at 6.1 m/s toward C, where he catches the ball. The ball is thrown at an angle of 45 above the horizontal, and it is thrown and caught at the same height above the ground. Determine the magnitude of the ball s initial velocit and the length of time it is in the air. Set as the horizontal motion of the football, as the vertical motion of the football and z as the horizontal motion of the receiver. Set g = 9.81 m/s, θ = 45. We have C Receiver s path Path of the ball a z =, v z = 6.1 m/s,s z = ( 6.1 m/s)t a = g, v = gt + v sin θ, s = 1 gt + v sin θ t 9 a =, v = v cos θ, s = v cos θ t When the ball is caught we have s z = ( 6.1 m/s)t B 9.1 m A = 1 gt + v sin θ t s = v cos θ t s = s z + ( 9.1 m) We can solve these four equations for the four unknowns s,s z,v,t We find t = 1.67 s, v = 11.6 m/s 44