2 MOTION ALONG A STRAIGHT LINE

Transcription

1 MOTION ALONG A STRAIGHT LINE Download full Solution manual for Universit phsics with modern phsics 14t IDENTIFY: = v t SET UP: We know the average velocit is 6.5 m/s. =v EXECUTE: t = 5. m EVALUATE: In round numbers, 6 m/s 4 s = 4 m 5 m, so the answer is reasonable... IDENTIFY: v = t SET UP: das = s. At the release point, = m. EXECUTE: (a) v = = m 1 = 4.4 m/s. t s (b) For the round trip, = 1 and =. The average velocit is zero. EVALUATE: The average velocit for the trip from the nest to the release point is positive..3. IDENTIFY: Target variable is the time t it takes to make the trip in heav traffic. Use Eq. (.) that relates the average velocit to the displacement and average time. SET UP: v = so = v t and t =. v t EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two cities, where the time is 1 h and 5 min, which is 11 min: =v t = (15 km/h)(1 h/6 min)(11 min) = 19.5 km. Now use v for heav traffic to calculate t; is the same as before: t = = 19.5 km =.75 h = h and 45 min. v 7 km/h The additional time is ( h and 45 min) (1 h and 5 min) = (1 h and 15 min) (1 h and 5 min) = 55 min. EVALUATE: At the normal speed of 15 km/s the trip takes 11 min, but at the reduced speed of 7 km/h it takes 165 min. So decreasing our average speed b about 3% adds 55 min to the time, which is 5% of 11 min. Thus a 3% reduction in speed leads to a 5% increase in travel time. This result (perhaps surprising) occurs because the time interval is inversel proportional to the average speed, not directl proportional to it..4. IDENTIFY: The average velocit is v =. Use the average speed for each segment to find the time t traveled in that segment. The average speed is the distance traveled divided b the time. SET UP: The post is 8 m west of the pillar. The total distance traveled is m + 8 m = 48 m. EXECUTE: (a) The eastward run takes time m = 4. s and the westward run takes 5. m/s 8 m = 7. s. The average speed for the entire trip is 48 m = 4.4 m/s. 4. m/s 11. s (b) v = = 8 m =.73 m/s. The average velocit is directed westward. t 11. s

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3 - Chapter EVALUATE: The displacement is much less than the distance traveled, and the magnitude of the average velocit is much less than the average speed. The average speed for the entire trip has a value that lies between the average speed for the two segments..5. IDENTIFY: Given two displacements, we want the average velocit and the average speed. SET UP: The average velocit is v = and the average speed is just the total distance walked divided b the total time to walk this distance. t EXECUTE: (a) Let + be east. = 6. m 4. m =. m and t = 8. s s = 64. s. So v = =. m =.31 m/s. t 64. s (b) average speed = 6. m + 4. m = 1.56 m/s 64. s EVALUATE: The average speed is much greater than the average velocit because the total distance walked is much greater than the magnitude of the displacement vector..6. IDENTIFY: The average velocit is v =. Use (t) to find for each t. t SET UP: () =, (. s) = 5.6 m, and (4. s) =.8 m EXECUTE: (a) v (b) v = 5. 6 m = +.8 m/s. s =. 8 m = + 5. m/s 4. s (c) v =. 8 m 5.6 m = m/s. s EVALUATE: The average velocit depends on the time interval being considered..7. (a) IDENTIFY: Calculate the average velocit using v =. t SET UP: v = so use (t) to find the displacement for this time interval. t EXECUTE: t = : = t = 1. s: = (. 4 m/s )(1. s) (. 1 m/s 3 )(1. s) 3 = 4 m 1 m = 1 m. Then v = = 1 m = 1. m/s. t 1. s (b) IDENTIFY: Use v = d to calculate v (t) and evaluate this epression at each specified t. SET UP: v = d = bt 3ct. EXECUTE: (i) t = : v = (ii) t = 5. s: v = (. 4 m/s )(5. s) 3(. 1 m/s 3 )(5. s) = 4. m/s 9. m/s = 15. m/s. (iii) t = 1. s: v = (. 4 m/s )(1. s) 3(. 1 m/s 3 )(1. s) = 48. m/s 36. m/s = 1. m/s. (c) IDENTIFY: Find the value of t when v (t) SET UP: v = bt 3ct v = at t =. v = net when bt 3ct = = = b = (.4 m/s ) EXECUTE: b 3ct so t 3c 3(.1 m/s 3 ) = 13.3 s from part (b) is zero. EVALUATE: v (t) for this motion sas the car starts from rest, speeds up, and then slows down again. Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

4 Motion Along a Straight Line IDENTIFY: We know the position (t) of the bird as a function of time and want to find its instantaneous velocit at a particular time. d d 3 3 )t SET UP: The instantaneous velocit is v (t) = = 8. m + (1. 4 m/s)t (.45 m/s. EXECUTE: v (t ) = d = 1. 4 m/s (.135 m/s 3 )t. Evaluating this at t = 8. s gives v = 3.76 m/s. EVALUATE: The acceleration is not constant in this case..9. IDENTIFY: The average velocit is given b v =. We can find the displacement t for each t constant velocit time interval. The average speed is the distance traveled divided b the time. SET UP: For t = to t =. s, v =. m/s. For t =. s to t = 3. s, v = 3. m/s. In part (b), v = 3. m/s for t =. s to t = 3. s. When the velocit is constant, = v t. EXECUTE: (a) For t = to t =. s, = (. m/s)(. s) = 4. m. For t =. s to t = 3. s, = (3. m/s)(1. s) = 3. m. For the first 3. s, = 4. m + 3. m = 7. m. The distance traveled is also 7. m. The average velocit is v = = 7. m =.33 m/s. The average speed is also.33 m/s. t 3. s (b) For t =. s to 3. s, = ( 3. m/s)(1. s) = 3. m. For the first 3. s, = 4. m + ( 3. m) = + 1. m. The ball travels 4. m in the +-direction and then 3. m in the -direction, so the distance traveled is still 7. m. v = = 1. m =.33 m/s. The average speed is t 3. s 7. m =. 33 m/s. 3. s EVALUATE: When the motion is alwas in the same direction, the displacement and the distance traveled are equal and the average velocit has the same magnitude as the average speed. When the motion changes direction during the time interval, those quantities are different..1. IDENTIFY and SET UP: The instantaneous velocit is the slope of the tangent to the versus t graph. EXECUTE: (a) The velocit is zero where the graph is horizontal; point IV. (b) The velocit is constant and positive where the graph is a straight line with positive slope; point I. (c) The velocit is constant and negative where the graph is a straight line with negative slope; point V. (d) The slope is positive and increasing at point II. (e) The slope is positive and decreasing at point III. EVALUATE: The sign of the velocit indicates its direction..11. IDENTIFY: Find the instantaneous velocit of a car using a graph of its position as a function of time. SET UP: The instantaneous velocit at an point is the slope of the versus t graph at that point. Estimate the slope from the graph. EXECUTE: A: v = 6.7 m/s; B: v = 6.7 m/s; C: v = ; D: v = 4. m/s; E: v = 4. m/s; F: v = 4. m/s; G: v =. EVALUATE: The sign of v shows the direction the car is moving. v is constant when versus t is a straight line..1. IDENTIFY: a =. a (t) is the slope of the v v t SET UP: 6 km/h = 16.7 m/s versus t graph. EXECUTE: (a) (i) a = m/s = 1.7 m/s. (ii) a = 16.7 m/s = 1.7 m/s. 1 s 1 s (iii) v = and a =. (iv) v = and a =. v (b) At t = s, is constant and a =. At t = 35 s, the graph of v versus t is a straight line and a = a = 1.7 m/s. Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

5 -4 Chapter EVALUATE: When a and v have the same sign the speed is increasing. When the have opposite signs, the speed is decreasing..13. IDENTIFY: The average acceleration for a time interval t is given b a =. t SET UP: Assume the car is moving in the + direction. 1 mi/h 447 m/s, so 6 mi/h 6 8 m/s, mi/h = 89.4 m/s and 53 mi/h = m/s. EXECUTE: (a) The graph of v versus t is sketched in Figure.13. The graph is not a straight line, so the acceleration is not constant. (b) (i) a = 6. 8 m/s = 1.8 m/s (ii) a = 89.4 m/s 6.8 m/s = 3.5 m/s.1 s. s.1 s (iii) a = 113.1m/s 89.4 m/s =.718 m/s. The slope of the graph of v versus t decreases as t 53 s. s increases. This is consistent with an average acceleration that decreases in magnitude during each successive time interval. EVALUATE: The average acceleration depends on the chosen time interval. For the interval between and 53 s, a = 113.1m/s =.13 m/s. 53 s v Figure IDENTIFY: We know the velocit v(t) of the car as a function of time and want to find its acceleration at the instant that its velocit is 1. m/s. 3 dv 3 )t a (t) = d (.86 m/s SET UP: We know that v (t) = (.86 m/s )t and that =. dv 3 3 EXECUTE: a (t ) = =(1.7 m/s )t. When v = 1. m/s, (.86 m/s )t = 1. m/s, which gives t = s. At this time, a = 6.4 m/s. EVALUATE: The acceleration of this car is not constant..15. IDENTIFY and SET UP: Use v = d and a = dv to calculate v (t) and a (t). EXECUTE: v = d =. cm/s (.15 cm/s )t a = dv =.15 cm/s (a) At t =, = 5. cm, v =. cm/s, a =.15 cm/s. (b) Set v = and solve for t: t = 16. s. (c) Set = 5. cm and solve for t. This gives t = and t = 3. s. The turtle returns to the starting point after 3. s. (d) The turtle is 1. cm from starting point when = 6. cm or = 4. cm. Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

6 Motion Along a Straight Line -5 Set = 6. cm and solve for t: t = 6. s and t = 5. 8 s. At t = 6. s, v = cm/s. At t = 5.8 s, v = 1. 3 cm/s. Set = 4. cm and solve for t: t = 36.4 s (other root to the quadratic equation is negative and hence nonphsical). At t = 36.4 s, v =. 55 cm/s. (e) The graphs are sketched in Figure.15. Figure.15 EVALUATE: The acceleration is constant and negative. v is linear in time. It is initiall positive, decreases to zero, and then becomes negative with increasing magnitude. The turtle initiall moves farther awa from the origin but then stops and moves in the -direction..16. IDENTIFY: Use a = v t, with t = 1 s in all cases. SET UP: v is negative if the motion is to the left. EXECUTE: (a) [(5. m/s) (15. m/s)]/(1 s) = 1. m/s (b) [( 15. m/s) ( 5. m/s)]/(1 s) = 1. m/s (c) [( 15. m/s) (+ 15. m/s)]/(1 s) = 3. m/s EVALUATE: In all cases, the negative acceleration indicates an acceleration to the left..17. IDENTIFY: The average acceleration is a = v. Use v (t) to find v t acceleration is a = dv. at each t. The instantaneous SET UP: v () = 3. m/s and v (5. s) = 5.5 m/s. EXECUTE: (a) a = v = 5. 5 m/s 3. m/s =.5 m/s t 5. s dv (b) a = = (. 1 m/s 3 )(t ) = (. m/s 3 )t. At t =, a =. At t = 5. s, a = 1. m/s. (c) Graphs of v (t) and a (t) are given in Figure.17 (net page). EVALUATE: a (t) is the slope of v (t) and increases as t increases. The average acceleration for t = to t = 5. s equals the instantaneous acceleration at the midpoint of the time interval, t =.5 s, since a (t) is a linear function of t. Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

7 -6 Chapter Figure IDENTIFY: v (t) = d and a (t) = dv d SET UP: (t n ) = nt n 1 for n 1. EXECUTE: (a) v (t ) = (9. 6 m/s )t (.6 m/s 6 )t 5 and a (t ) = 9. 6 m/s (3. m/s 6 )t 4. Setting v = gives t = and t =. s. At t, =.17 m and a 9 6 m/s. At t s, = 15. m and a = 38.4 m/s. (b) The graphs are given in Figure.18. EVALUATE: For the entire time interval from t = to t =. s, the velocit v is positive and increases. While a is also positive the speed increases and while a is negative the speed decreases. Figure IDENTIFY: Use the constant acceleration equations to find v and a. (a) SET UP: The situation is sketched in Figure.19. Figure.19 = 7. m t = 6. s v = 15. m/s v =? Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

8 Motion Along a Straight Line -7 (b) Use v v EXECUTE: Use = + v t, so v = ( ) v = (7. m) 15. m/s = 8.33 m/s. t 6. s = v + a t, so a = v v = 15. m/s 5. m/s = m/s. t 6. s EVALUATE: The average velocit is (7. m)/(6. s) = 11.7 m/s. The final velocit is larger than this, so the antelope must be speeding up during the time interval; v < v and a >... IDENTIFY: In (a) find the time to reach the speed of sound with an acceleration of 5g, and in (b) find his speed at the end of 5. s if he has an acceleration of 5g. SET UP: Let + be in his direction of motion and assume constant acceleration of 5g so the standard kinematics equations appl so v = v + a t. (a) v = 3(331 m/s) = 993 m/s, v =, and a = 5g = 49. m/s. (b) t = 5. s EXECUTE: (a) v = v + a t and t = v v than 5. s. (b) v a = v + a t = + (49. m/s )(5. s) = 45 m/s. = 993 m/s =.3 s. Yes, the time required is larger 49. m/s EVALUATE: In 5. s he can onl reach about /3 the speed of sound without blacking out..1. IDENTIFY: For constant acceleration, the standard kinematics equations appl. SET UP: Assume the ball starts from rest and moves in the + -direction. EXECUTE: a = (a) = 1.5 m, v = 45. m/s and v =. v = v + a ( ) gives v v (45. m/s) ( = = 675 m/s ) (1.5 m). (b) = v + v t gives t = ( ) = (1.5 m) =.667 s v + v 45. m/s EVALUATE: We could also use v = v + a t to find t = v = 45. m/s =.667 s which agrees with a 675 m/s our previous result. The acceleration of the ball is ver large... IDENTIFY: For constant acceleration, the standard kinematics equations appl. SET UP: Assume the ball moves in the + direction. EXECUTE: (a) v = m/s, v = and t = 3. ms. v = v + a t gives a = v v = m/s = 44 m/s. t s (b) = v + v t = m/s ( s) = 1.1 m. EVALUATE: We could also use = v t + 1 a t to calculate : = 1 (44 m/s )( s) = 1.1 m, which agrees with our previous result. The acceleration of the ball is ver large..3. IDENTIFY: Assume that the acceleration is constant and appl the constant acceleration kinematic equations. Set a equal to its maimum allowed value. SET UP: Let + be the direction of the initial velocit of the car. a = 5 m/s. 15 km/h = 9.17 m/s. EXECUTE: v = 9.17 m/s. v =. v = v + a ( ) gives = v v = (9.17 m/s) = 1.7 m. a ( 5 m/s ) EVALUATE: The car frame stops over a shorter distance and has a larger magnitude of acceleration. Part of our 1.7 m stopping distance is the stopping distance of the car and part is how far ou move relative to the car while stopping. Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

9 -8 Chapter.4. IDENTIFY: In (a) we want the time to reach Mach 4 with an acceleration of 4g, and in (b) we want to know how far he can travel if he maintains this acceleration during this time. SET UP: Let + be the direction the jet travels and take =. With constant acceleration, the equations = v + a t and = + v t + 1 a t both appl. a = 4g = 39. m/s, v = 4(331 m/s) = 134 m/s, and v v =. EXECUTE: (a) Solving v = v + a t for t gives t = v v = 134 m/s = 33.8 s. a 39. m/s (b) = + v t + 1 a t = 1 (39. m/s )(33. 8 s) = m =.4 km. EVALUATE: The answer in (a) is about ½ min, so if he wanted to reach Mach 4 an sooner than that, he would be in danger of blacking out..5. IDENTIFY: If a person comes to a stop in 36 ms while slowing down with an acceleration of 6g, how far does he travel during this time? SET UP: Let + be the direction the person travels. v = (he stops), a is negative since it is opposite to the direction of the motion, and t = 36 ms = s. The equations v = + v t + 1 a t both appl since the acceleration is constant. = v + a t and EXECUTE: Solving v = v + a t for v gives v = a t. Then = + v t + 1 a t gives = 1 a t = 1 ( 588 m/s )( s) = 38 cm. EVALUATE: Notice that we were not given the initial speed, but we could find it: v = a t = ( 588 m/s )( s) = 1 m/s = 47 mph..6. IDENTIFY: In (a) the hip pad must reduce the person s speed from. m/s to 1.3 m/s over a distance of. cm, and we want the acceleration over this distance, assuming constant acceleration. In (b) we want to find out how long the acceleration in (a) lasts. SET UP: Let + be downward. v =. m/s, v = 1.3 m/s, and =. m. The equations v + v v = v + a ( ) and = t appl for constant acceleration. EXECUTE: (a) Solving v = v + a ( ) for a gives v v (1. 3 m/s) (. m/s) a = ( = = 58 m/s = 5.9g. ) (. m) v + v ( ) (. m) (b) = t gives t = = = 1 ms. v +v. m/s m/s EVALUATE: The acceleration is ver large, but it onl lasts for 1 ms so it produces a small velocit change..7. IDENTIFY: We know the initial and final velocities of the object, and the distance over which the velocit change occurs. From this we want to find the magnitude and duration of the acceleration of the object. SET UP: The constant-acceleration kinematics formulas appl. v = v + a ( ), where v =, v = m/s, and = 4. m. EXECUTE: (a) v = v v v ( m/s) + a ( ) gives a = m/s = = = g. ( ) (4. m) (b) v = v + a t gives t = v v = m/s = 1.6 ms. a m/s EVALUATE: (c) The calculated a is less than 45, g so the acceleration required doesn t rule out this hpothesis..8. IDENTIFY: Appl constant acceleration equations to the motion of the car. SET UP: Let + be the direction the car is moving. EXECUTE: (a) From v = v + a ( ), with v =, a = v = ( m/s) = 1.67 m/s. ( ) (1 m)

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11 Motion Along a Straight Line -9 (b) Using Eq. (.14), t = ( )/v = (1 m)/( m/s) = 1 s. (c) (1 s)( m/s) = 4 m. EVALUATE: The average velocit of the car is half the constant speed of the traffic, so the traffic travels twice as far..9. IDENTIFY: The average acceleration is a = v. For constant acceleration, the standard kinematics t equations appl. SET UP: Assume the rocket ship travels in the + direction. 161 km/h = 44.7 m/s and 161 km/h = 447. m/s. 1. min = 6. s EXECUTE: (a) (i) a = v = 44.7 m/s = 5.59 m/s t 8. s (ii) a = 447. m/s 44.7 m/s = 7.74 m/s 6. s 8. s (b) (i) t = 8. s, v =, and v = 44.7 m/s. = v + v t = m/s (8. s) = 179 m. (ii) t = 6. s 8. s = 5. s, v = 44.7 m/s, and v = 447. m/s. = v + v t = 44.7 m/s m/s(5. s) = m. EVALUATE: When the acceleration is constant the instantaneous acceleration throughout the time interval equals the average acceleration for that time interval. We could have calculated the distance in part (a) as = v t + 1 a t = 1 (5.59 m/s )(8. s) = 179 m, which agrees with our previous calculation..3. IDENTIFY: The acceleration a is the slope of the graph of v versus t. SET UP: The signs of v and of a indicate their directions. EXECUTE: (a) Reading from the graph, at t = 4. s, v =.7 cm/s, to the right and at t = 7. s, v = 1.3 cm/s, to the left. (b) v versus t is a straight line with slope 8. cm/s = 1.3 cm/s. The acceleration is constant and 6. s equal to 1.3 cm/s, to the left. It has this value at all times. (c) Since the acceleration is constant, = v t + 1 a t. For t = to 4.5 s, = (8. cm/s)(4.5 s) + 1 ( 1.3 cm/s )(4.5 s) =.8 cm. For t = to 7.5 s, = (8. cm/s)(7.5 s) + 1 ( 1.3 cm/s )(7.5 s) = 3.4 cm (d) The graphs of a and versus t are given in Figure.3. EVALUATE: In part (c) we could have instead used = v + v t. Figure.3 Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

12 -1 Chapter.31. (a) IDENTIFY and SET UP: The acceleration a at time t is the slope of the tangent to the v versus t curve at time t. EXECUTE: At t = 3 s, the v versus t curve is a horizontal straight line, with zero slope. Thus a =. At t = 7 s, the v versus t curve is a straight-line segment with slope 45 m/s m/s = 6. 3 m/s. 9 s 5 s Thus a = 6. 3 m/s. At t = 11 s the curve is again a straight-line segment, now with slope 45 m/s = 11. m/s. 13 s 9 s Thus a = 11. m/s. EVALUATE: a = when v is constant, a > when v is positive and the speed is increasing, and a < when v is positive and the speed is decreasing. (b) IDENTIFY: Calculate the displacement during the specified time interval. SET UP: We can use the constant acceleration equations onl for time intervals during which the acceleration is constant. If necessar, break the motion up into constant acceleration segments and appl the constant acceleration equations for each segment. For the time interval t = to t = 5 s the acceleration is constant and equal to zero. For the time interval t = 5 s to t = 9 s the acceleration is constant and equal to 6. 5 m/s. For the interval t = 9 s to t = 13 s the acceleration is constant and equal to 11. m/s. EXECUTE: During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas can be used. v = m/s a = t = 5 s =? = v t (a = so no 1 a t term) = ( m/s)(5 s) = 1 m; this is the distance the officer travels in the first 5 seconds. During the interval t = 5 s to 9 s the acceleration is again constant. The constant acceleration formulas can be applied to this 4-second interval. It is convenient to restart our clock so the interval starts at time t = and ends at time t = 4 s. (Note that the acceleration is not constant over the entire t = to t = 9 s interval.) v = m/s a = 6.5 m/s t = 4 s = 1 m =? = v t + 1 a t = ( m/s)(4 s) + 1 (6. 5 m/s )(4 s) = 8 m + 5 m = 13 m. Thus + 13 m = 1 m + 13 m = 3 m. At t = 9 s the officer is at = 3 m, so she has traveled 3 m in the first 9 seconds. During the interval t = 9 s to t = 13 s the acceleration is again constant. The constant acceleration formulas can be applied for this 4-second interval but not for the whole t = to t = 13 s interval. To use the equations restart our clock so this interval begins at time t = and ends at time t = 4 s. v = 45 m/s (at the start of this time interval) a = 11. m/s t = 4 s = 3 m =? = v t + 1 a t 1 = (45 m/s)(4 s) + ( 11. m/s )(4 s) = 18 m m = 9.4 m. Thus = m = 3 m m = 3 m. At t = 13 s the officer is at = 3 m, so she has traveled 3 m in the first 13 seconds. EVALUATE: The velocit v is alwas positive so the displacement is alwas positive and displacement and distance traveled are the same. The average velocit for time interval t is v = / t. For t = to 5 s, v = m/s. For t = to 9 s, v = 6 m/s. For t = to 13 s, v = 5 m/s. These results are consistent with the figure in the tetbook..3. IDENTIFY: v (t) is the slope of the versus t graph. Car B moves with constant speed and zero acceleration. Car A moves with positive acceleration; assume the acceleration is constant. Copright 16 Pearson Education, Inc. 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.

13 Motion Along a Straight Line -11 SET UP: For car B, v is positive and a =. For car A, a is positive and v increases with t. EXECUTE: (a) The motion diagrams for the cars are given in Figure.3a. (b) The two cars have the same position at times when their -t graphs cross. The figure in the problem shows this occurs at approimatel t = 1 s and t = 3 s. (c) The graphs of v versus t for each car are sketched in Figure.3b. (d) The cars have the same velocit when their -t graphs have the same slope. This occurs at approimatel t = s. (e) Car A passes car B when A moves above B in the -t graph. This happens at t = 3 s. (f) Car B passes car A when B moves above A in the -t graph. This happens at t = 1 s. EVALUATE: When a =, the graph of v versus t is a horizontal line. When a is positive, the graph of v versus t is a straight line with positive slope. Figure IDENTIFY: For constant acceleration, the kinematics formulas appl. We can use the total displacement ande final velocit to calculate the acceleration and then use the acceleration and shorter distance to find the speed. SET UP: Take + to be down the incline, so the motion is in the + direction. The formula v = v + a( ) applies. EXECUTE: First look at the motion over 6.8 m. We use the following numbers: v =, = 6.8 m, and v = 3.8 /s. Solving the above equation for a gives a = 1.6 m/s. Now look at the motion over the 3.4 m using v =, a = 1.6 m/s and = 3.4 m. Solving the same equation, but this time for v, gives v =.69 m/s. EVALUATE: Even though the block has traveled half wa down the incline, its speed is not half of its speed at the bottom..34. IDENTIFY: Appl the constant acceleration equations to the motion of each vehicle. The truck passes the car when the are at the same at the same t >. SET UP: The truck has a =. The car has v =. Let + be in the direction of motion of the vehicles. Both vehicles start at =. The car has a C =.8 m/s. The truck has v =. m/s. EXECUTE: (a) = v t + 1 a t gives = v t and = 1 a t. Setting = gives t = and T T C C T C v = 1 a t, so t = v T = (. m/s) = 14.9 s. At this t, = (. m/s)(14.9 s) = 86 m and T C a.8 m/s T C = 1 (3. m/s )(14.9 s) = 86 m. The car and truck have each traveled 86 m. (b) At t = 14.9 s, the car has v = v + a t = (.8 m/s )(14.9 s) = 4 m/s. (c) T = v Tt and C 1 a Ct. The -t graph of the motion for each vehicle is sketched in Figure.34a. (d) v T = v T. v C = a Ct. The v -t graph for each vehicle is sketched in Figure.34b (net page). EVALUATE: When the car overtakes the truck its speed is twice that of the truck. Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

14 -1 Chapter Figure IDENTIFY: Appl the constant acceleration equations to the motion of the flea. After the flea leaves the ground, a = g, downward. Take the origin at the ground and the positive direction to be upward. (a) SET UP: At the maimum height v v = =.44 m a = 9.8 m/s v =? v = v + a ( ) EXECUTE: v = a ( ) = ( 9. 8 m/s )(. 44 m) =.94 m/s (b) SET UP: When the flea has returned to the ground =. = v = +.94 m/s a = 9.8 m/s t =? = v t + 1 a t EXECUTE: With = this gives t = v = (.94 m/s) =. 6 s. a 9.8 m/s EVALUATE: We can use v v a t to show that with v 94 m/s, v after.3 s..36. IDENTIFY: The rock has a constant downward acceleration of 9.8 m/s. We know its initial velocit and position and its final position. SET UP: We can use the kinematics formulas for constant acceleration. EXECUTE: (a) = 3 m, v =. m/s, a = 9.8 m/s. The kinematics formulas give v = v + a ( ) = (. m/s) + ( 9. 8 m/s )( 3 m) = 3.74 m/s, so the speed is 3.7 m/s. (b) v v a t and t = v v = 3.74 m/s. m/s = 5.59 s. a 9.8 m/s EVALUATE: The vertical velocit in part (a) is negative because the rock is moving downward, but the speed is alwas positive. The 5.59 s is the total time in the air..37. IDENTIFY: The pin has a constant downward acceleration of 9.8 m/s and returns to its initial position. SET UP: We can use the kinematics formulas for constant acceleration. EXECUTE: The kinematics formulas give = v t + 1 a t. We know that =, so v t = = (8. m/s) = s. a 9.8 m/s EVALUATE: It takes the pin half this time to reach its highest point and the remainder of the time to return..38. IDENTIFY: The putt has a constant downward acceleration of 9.8 m/s. We know the initial velocit of the putt and the distance it travels. Copright 16 Pearson Education, Inc. 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.

15 Motion Along a Straight Line -13 SET UP: We can use the kinematics formulas for constant acceleration. EXECUTE: (a) v = 9.5 m/s and = 3.6 m, which gives = v + a ( ) = (9. 5 m/s) + ( 9. 8 m/s )(3. 6 m) = 4.44 m/s v (b) t = v v = m/s 9.5 m/s =.517 s a 9.8 m/s EVALUATE: The putt is stopped b the ceiling, not b gravit..39. IDENTIFY: A ball on Mars that is hit directl upward returns to the same level in 8.5 s with a constant downward acceleration of.379g. How high did it go and how fast was it initiall traveling upward? SET UP: Take + upward. v at the maimum height. a =. 379g = m/s. The constant- acceleration formulas v = v + a t and = + v t + 1 a t both appl. EXECUTE: Consider the motion from the maimum height back to the initial level. For this motion v = and t = 4. 5 s. = + v t + 1 a t = 1 ( m/s )(4. 5 s) = 33.5 m. The ball went 33.5 m above its original position. (b) Consider the motion from just after it was hit to the maimum height. For this motion v = and t = 4. 5 s. v v a t gives v = a t = ( m/s )(4. 5 s) = m/s. (c) The graphs are sketched in Figure.39. Figure.39 EVALUATE: The answers can be checked several was. For eample, v =, v = 15.8 m/s, and v v (15.8 m/s) a = 3.71 m/s in v = v + a ( ) gives = = = 33.6 m, which a ( 3.71 m/s ) agrees with the height calculated in (a)..4. IDENTIFY: Appl constant acceleration equations to the motion of the lander. SET UP: Let + be downward. Since the lander is in free-fall, a = m/s. EXECUTE: v 8 m/s, = 5. m, a = m/s in v = v + a v ( ) gives = v + a ( ) = (. 8 m/s) + (1. 6 m/s )(5. m) = 4.1 m/s. EVALUATE: The same descent on earth would result in a final speed of 9.9 m/s, since the acceleration due to gravit on earth is much larger than on the moon..41. IDENTIFY: Appl constant acceleration equations to the motion of the meterstick. The time the meterstick falls is our reaction time. SET UP: Let + be downward. The meter stick has v = and a = 9.8 m/s. Let d be the distance the meterstick falls. EXECUTE: (a) = v t + 1 a t gives d = (4.9 m/s )t and t = d. 4.9 m/s (b) t =.176 m =.19 s 4.9 m/s EVALUATE: The reaction time is proportional to the square of the distance the stick falls. Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

16 -14 Chapter.4. IDENTIFY: Appl constant acceleration equations to the vertical motion of the brick. SET UP: Let + be downward. a = 9.8 m/s EXECUTE: (a) v =, t = 1.9 s, a = 9.8 m/s. = v t + 1 a t = 1 (9. 8 m/s )(1.9 s) = 17.7 m. The building is 17.7 m tall. (b) v = v + a t = + (9. 8 m/s )(1.9 s) = 18.6 m/s (c) The graphs of a, v and versus t are given in Figure.4. Take = at the ground. v + v EVALUATE: We could use either = t or v = v + a ( ) to check our results. Figure IDENTIFY: When the onl force is gravit the acceleration is 9.8 m/s, downward. There are two intervals of constant acceleration and the constant acceleration equations appl during each of these intervals. SET UP: Let + be upward. Let = at the launch pad. The final velocit for the first phase of the motion is the initial velocit for the free-fall phase. EXECUTE: (a) Find the velocit when the engines cut off. = 55 m, a =.5 m/s, v =. v = v + a ( ) gives v = (. 5 m/s )(55 m) = 48.6 m/s. Now consider the motion from engine cut-off to maimum height: = 55 m, v = m/s, v = (at the maimum height), a = 9.8 m/s. v = v + a ( ) gives = v v = (48.6 m/s) =11 m and = 11 m + 55 m = 646 m. a ( 9.8 m/s ) (b) Consider the motion from engine failure until just before the rocket strikes the ground: = 55 m, a = 9.8 m/s,v = m/s. v = v + a ( ) gives v = (48. 6 m/s) + ( 9. 8 m/s )( 55 m) = 11 m/s. Then v = v + a t gives t = v v = 11 m/s 48.6 m/s = 16.4 s. a 9.8 m/s (c) Find the time from blast-off until engine failure: = 55 m, v =, a = +.5 m/s. = v t + 1 a t gives t = ( ) = (55 m) = 1.6 s. The rocket strikes the launch pad a.5 m/s 1. 6 s s = 38. s after blast-off. The acceleration a is +.5 m/s from t = to t = 1.6 s. It is 9.8 m/s from t = 1.6 s to 38. s. v = v + a t applies during each constant acceleration segment, so the graph of v versus t is a straight line with positive slope of.5 m/s during the blast-off phase and with negative slope of 9.8 m/s after engine failure. During each phase = v t + 1 a t. The sign of a determines the curvature of (t). At t = 38. s the rocket has returned to are sketched in Figure.43. =. The graphs Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

17 Motion Along a Straight Line -15 EVALUATE: In part (b) we could have found the time from = v t + 1 a t, finding v first allows us to avoid solving for t from a quadratic equation. Figure IDENTIFY: Appl constant acceleration equations to the vertical motion of the sandbag. SET UP: Take + upward. a = 9.8 m/s. The initial velocit of the sandbag equals the velocit of the balloon, so v = + 5. m/s. When the balloon reaches the ground, maimum height the sandbag has v =. = 4. m. At its EXECUTE: (a) t =.5 s: = v t + 1 a t = (5. m/s)(. 5 s) + 1 ( 9. 8 m/s )(. 5 s) =.94 m. The sandbag is 4.9 m above the ground. v = v + a t = + 5. m/s + ( 9. 8 m/s )(. 5 s) =.55 m/s. t = 1. s: the ground. v = (5. m/s)(1. s) + 1 ( 9. 8 m/s )(1. s) =.1 m. The sandbag is 4.1 m above = v + a t = + 5. m/s + ( 9. 8 m/s )(1. s) = 4.8 m/s. = 5. m/s, a (b) = 4. m, v = 9.8 m/s. = v t + 1 a t gives 4. m = (5. m/s)t (4.9 m/s )t. (4. 9 m/s )t (5. m/s)t 4. m = and 1 t = 9.8 (c) v (d) v 5. ± ( 5. ) 4(4. 9)( 4. ) s = (. 51±.9) s. t must be positive, so t = 3.41 s. = v + a t = + 5. m/s + ( 9. 8 m/s )(3. 41 s) = 8.4 m/s = 5. m/s, a = 9.8 m/s, v =. v = v + a ( ) gives = v v = (5. m/s) = 1.8 m. The maimum height is 41.3 m above the ground. a ( 9.8 m/s ) (e) The graphs of a, v, and versus t are given in Figure.44. Take = at the ground. EVALUATE: The sandbag initiall travels upward with decreasing velocit and then moves downward with increasing speed. Figure.44 Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

18 -16 Chapter.45. IDENTIFY: Use the constant acceleration equations to calculate a and. (a) SET UP: v = 4 m/s, v =, t =.9 s, a =? v = v + a t EXECUTE: a = v v = 4 m/s = 49 m/s t.9 s (b) a /g = (49 m/s )/(9. 8 m/s ) = 5.4 (c) = v t + 1 a t = + 1 (49 m/s )(. 9 s) = 11 m (d) SET UP: Calculate the acceleration, assuming it is constant: t = 1.4 s, v = 83 m/s, v = (stops), a =? v = v + a t EXECUTE: a = v v = 83 m/s = m/s t 1.4 s a /g = ( m/s )/(9.8 m/s ) =.6; a =.6g If the acceleration while the sled is stopping is constant then the magnitude of the acceleration is onl.6g. But if the acceleration is not constant it is certainl possible that at some point the instantaneous acceleration could be as large as 4g. EVALUATE: It is reasonable that for this motion the acceleration is much larger than g..46. IDENTIFY: Since air resistance is ignored, the egg is in free-fall and has a constant downward acceleration of magnitude 9.8 m/s. Appl the constant acceleration equations to the motion of the egg. SET UP: Take + to be upward. At the maimum height, v =. EXECUTE: (a) = 3. m, t = 5. s, a = 9.8 m/s. = v t + 1 a t gives v = 1 a t = 3. m 1 ( 9.8 m/s )(5. s) = m/s. t 5. s (b) v = m/s, v = (at the maimum height), a = 9.8 m/s. v = v + a ( ) gives = v v = (18.5 m/s) = 17.5 m. a ( 9.8 m/s ) (c) At the maimum height v =. (d) The acceleration is constant and equal to 9.8 m/s, downward, at all points in the motion, including at the maimum height. (e) The graphs are sketched in Figure.46. v v EVALUATE: The time for the egg to reach its maimum height is t = = 18.5 m/s = 1.89 s. The a 9.8 m/s egg has returned to the level of the cornice after 3.78 s and after 5. s it has traveled downward from the cornice for 1. s. Figure.46 Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

19 Motion Along a Straight Line IDENTIFY: We can avoid solving for the common height b considering the relation between height, time of fall, and acceleration due to gravit, and setting up a ratio involving time of fall and acceleration due to gravit. SET UP: Let g En be the acceleration due to gravit on Enceladus and let g be this quantit on earth. Let h be the common height from which the object is dropped. Let + be downward, so = h. v = EXECUTE: = v t + 1 a t gives h = 1 gt and h = 1 g t. Combining these two equations gives gt E = g En t En EVALUATE: E En En t E 1.75 s and gen = g = (9. 8 m/s ) =.868 m/s. t En 18.6 s The acceleration due to gravit is inversel proportional to the square of the time of fall..48. IDENTIFY: Since air resistance is ignored, the boulder is in free-fall and has a constant downward acceleration of magnitude 9.8 m/s. Appl the constant acceleration equations to the motion of the boulder. SET UP: Take + to be upward. EXECUTE: (a) v = + 4. m/s, v = +. m/s, a v v t = =. m/s 4. m/s = +.4 s. a 9.8 m/s (b) v = 9.8 m/s. v = v + at gives =. m/s. t = v v =. m/s 4. m/s = s. a 9.8 m/s (c) =, v = + 4. m/s, a = 9.8 m/s. = v t + 1 a t gives t = and t = v = (4. m/s) = s. a 9.8 m/s v v (d) v =, v = + 4. m/s, a. v = v + a t gives t = = 4. m/s = 9.8 m/s = 4.8 s. a 9.8 m/s (e) The acceleration is 9.8 m/s, downward, at all points in the motion. (f) The graphs are sketched in Figure.48. EVALUATE: v = at the maimum height. The time to reach the maimum height is half the total time in the air, so the answer in part (d) is half the answer in part (c). Also note that. 4 s < 4. 8 s < 6.1 s. The boulder is going upward until it reaches its maimum height and after the maimum height it is traveling downward. Figure IDENTIFY: The rock has a constant downward acceleration of 9.8 m/s. The constant-acceleration kinematics formulas appl. SET UP: The formulas = + v t + 1 a t and v = v + a ( ) both appl. Call + upward. First find the initial velocit and then the final speed. EXECUTE: (a) 6. s after it is thrown, the rock is back at its original height, so = at that instant. Using a = 9.8 m/s and t = 6. s, the equation = + v t + 1 a t gives v = 9.4 m/s. When the rock Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

20 -18 Chapter reaches the water, = 8. m. The equation v = v + a ( ) gives v = 37.6 m/s, so its speed is 37.6 m/s. EVALUATE: The final speed is greater than the initial speed because the rock accelerated on its wa down below the bridge..5. IDENTIFY: The acceleration is not constant, so we must use calculus instead of the standard kinematics formulas. SET UP: The general calculus formulas are v = v + t a and = + t v. First integrate a to find v(t), and then integrate that to find (t). EXECUTE: Find v(t): v (t ) = v + t a = v + t (.3 m/s 3 )(15. s t ). Carring out the integral and putting in the numbers gives v(t) = 8. m/s (.3 m/s 3 )[(15. s)t t /]. Now use this result to find (t). = + t v = t 8. m/s (.3 m/s 3 )((15. s)t t ), which gives = + (8. m/s)t (.3 m/s 3 )[(7.5 s)t t 3 /6)]. Using = 14. m and t = 1. s, we get = 47.3 m. EVALUATE: The standard kinematics formulas appl onl when the acceleration is constant..51. IDENTIFY: The acceleration is not constant, but we know how it varies with time. We can use the definitions of instantaneous velocit and position to find the rocket s position and speed. SET UP: The basic definitions of velocit and position are v (t ) = v + t a and = t v. EXECUTE: (a) v (t ) = t a = t (. 8 m/s 3 )t = (1.4 m/s 3 )t = t v = t (1. 4 m/s 3 )t = (.4667 m/s 3 )t 3. For t = 1. s, = 467 m. (b) = 35 m so ( m/s 3 )t 3 = 35 m and t = s. At this time v = (1. 4 m/s 3 )( s) = 11 m/s. EVALUATE: The time in part (b) is less than 1. s, so the given formulas are valid..5. IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use v = v + t a and = + t v. Use the values of v and of at t = 1. s to evaluate v and. SET UP: t n = 1 n + 1 tn 1, for n. EXECUTE: (a) v = v + t α t = v + 1 α t = v + (.6 m/s 3 )t. v = 5. m/s when t = 1. s gives v = 4.4 m/s. Then, at t =. s, v = 4. 4 m/s + (. 6 m/s 3 )(. s) = 6.8 m/s. (b) = + t (v + 1 α t ) = + v t α t 3. = 6. m at t = 1. s gives = 1.4 m. Then, at t =. s, = 1. 4 m + (4. 4 m/s)(. s) + 1 (1. m/s 3 )(. s) 3 = 11.8 m. 6 (c) (t) = 1. 4 m + (4. 4 m/s)t + (. m/s 3 )t 3. v (t) = 4. 4 m/s + (.6 m/s 3 )t. a (t) = (1. m/s 3 )t. The graphs are sketched in Figure.5. EVALUATE: We can verif that a = dv and v = d. Figure.5 Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

21 Motion Along a Straight Line (a) IDENTIFY: Integrate a (t) to find v (t) and then integrate v (t) to find (t ). SET UP: v = v + t a, a = At Bt with A = 1.5 m/s 3 and B =.1 m/s 4. EXECUTE: v = v + t (At Bt 1 ) = v + At rest at t = sas that v =, so At 1 Bt 3 3 v = 1 At 1 Bt 3 = 1 (1.5 m/s 3 )t 1 (.1 m/s 4 )t v = (.75 m/s 3 )t (.4 m/s 4 )t 3 SET UP: + t v EXECUTE: = + t ( 1 At 1 Bt 3 ) = + 1 At At the origin at t = sas that =, so = 1 At 3 1 Bt 4 = 1 (1.5 m/s 3 )t 3 1 (.1 m/s 4 )t = (.5 m/s 3 )t 3 (.1 m/s 4 )t 4 1 Bt 4 EVALUATE: We can check our results b using them to verif that v (t) = d and a (t) = dv. (b) IDENTIFY and SET UP: At time t, when v is a maimum, dv =. (Since a = dv, the maimum velocit is when a =. For earlier times a is positive so v negative and v is decreasing.) EXECUTE: a = dv = so At Bt = One root is t =, but at this time v = and not a maimum. The other root is t = A 1.5 m/s3 = = 1.5 s B.1 m/s 4 At this time v = (.75 m/s 3 )t (.4 m/s 4 )t 3 gives is still increasing. For later times a is v = (.75 m/s 3 )(1.5 s) (.4 m/s 4 )(1.5 s) 3 = 117. m/s 78.1 m/s = 39.1 m/s. EVALUATE: For t < 1.5 s, a > and v is increasing. For t > 1.5 s, a < and v is decreasing..54. IDENTIFY: a (t) is the slope of the v versus t graph and the distance traveled is the area under the v versus t graph. SET UP: The v versus t graph can be approimated b the graph sketched in Figure.54 (net page). EXECUTE: (a) Slope = a = for t 1.3 ms. (b) h = Area under v-t graph A + A 1 (1.3 ms)(133 cm/s) + (.5 ms 1.3 ms)(133 cm/s) ma Triangle Rectangle.5 cm 133 cm/s 5 (c) a = slope of v-t graph. a (.5 ms) a (1. ms) a (1.5 ms) = because the slope is zero. 1.3 ms = 1. 1 cm/s. (d) h = area under v-t graph. h(.5 ms) A Triangle = 1 (.5 ms)(33 cm/s) = cm. h (1. ms) A Triangle = 1 (1. ms)(1 cm/s) = 5. 1 cm. h(1.5 ms) A Triangle + A Rectangle = 1 (1.3 ms)(133 cm/s) + (. ms)(133 cm/s) =.11 cm. Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.

22 - Chapter EVALUATE: The acceleration is constant until t = 1.3 ms, and then it is zero. g = 98 cm/s. The acceleration during the first 1.3 ms is much larger than this and gravit can be neglected for the portion of the jump that we are considering. Figure IDENTIFY: The sprinter s acceleration is constant for the first. s but zero after that, so it is not constant over the entire race. We need to break up the race into segments. SET UP: When the acceleration is constant, the formula velocit is v =. t = v + v t applies. The average EXECUTE: (a) = v + v t = + 1. m/s (. s) = 1. m. (b) (i) 4. m at 1. m/s so time at constant speed is 4. s. The total time is 6. s, so v = = 5. m = 8.33 m/s. t 6. s v av - (ii i) v av - (ii) He runs 9. m at 1. m/s so the time at constant speed is 9. s. The total time is 11. s, so = 1 m =. 9 9 m/s. 11. s He runs 19 m at m/s. 1. m/s so time at constant speed is 19. s. His total time is 1. s, so = m 1. s EVALUATE: top speed..56. IDENTIFY: We know the vertical position of the lander as a function of time and want to use this to find its velocit initiall and just before it hits the lunar surface. SET UP: B definition, v (t) = d, so we can find v as a function of time and then evaluate it for the desired cases. EXECUTE: (a) v (t) = d = c +. At t =, v (t) = c = 6. m/s. The initial velocit is 6. m/s downward. (b) ( t) = sas b ct + =. The quadratic formula sas t = s ± 7.38 s. It reaches the surface at t = 1.19 s. At this time, v = 6. m/s + (1. 5 m/s )(1. 19 s) = 15.5 m/s. EVALUATE: The given formula for (t) is of the form = + vt + 1 at. For part (a), v = c = 6 m/s..57. IDENTIFY: In time t S the S-waves travel a distance d = v St S and in time t P the P-waves travel a distance d = v P t P. = 9.5 His average velocit keeps increasing because he is running more and more of the race at his SET UP: t S = t P + 33 s d d 1 1

23 EXECUTE: = + 33 s. d = 33 s and d = 5 km. v S vp 3.5 km/s 6.5 km/s Copright 16 Pearson Education, Inc. 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.

24 Motion Along a Straight Line -1 EVALUATE: The times of travel for each wave are t S = 71s and t P = 38 s..58. IDENTIFY: The brick has a constant downward acceleration, so we can use the usual kinematics formulas. We know that it falls 4. m in 1. s, but we do not know which second that is. We want to find out how far it falls in the net 1.-s interval. SET UP: Let the + direction be downward. The final velocit at the end of the first 1.-s interval will be the initial velocit for the second 1.-s interval. a = 9.8 m/s and the formula v t + 1 a t applies. EXECUTE: (a) First find the initial speed at the beginning of the first 1.-s interval. Appling the above formula with a = 9.8 m/s, t = 1. s, and = 4. m, we get v = 35.1 m/s. At the end of this 1.-s interval, the velocit is v = 35.1 m/s + (9.8 m/s )(1. s) = 44.9 m/s. This is v for the net 1.-s interval. Using v t + 1 a t with this initial velocit gives = 49.8 m. EVALUATE: The distance the brick falls during the second 1.-s interval is greater than during the first 1.-s interval, which it must be since the brick is accelerating downward..59. IDENTIFY: The average velocit is v =. t SET UP: Let + be upward. EXECUTE: (a) v (b) v = 1 m 63 m = 197 m/s 4.75 s = 1 m = 169 m/s 5.9 s EVALUATE: For the first 1.15 s of the flight, v constant the average velocit depends on the time = 63 m = 54.8 m/s. When the velocit isn t 1.15 s interval chosen. In this motion the velocit is increasing..6. IDENTIFY: Use constant acceleration equations to find for each segment of the motion. SET UP: Let + be the direction the train is traveling. EXECUTE: t = to 14. s: = v t + 1 a t = 1 (1.6 m/s )(14. s) = 157 m. At t = 14. s, the speed is v = v + a t = (1.6 m/s )(14. s) =.4 m/s. In the net 7. s, a = v t = (.4 m/s)(7. s) = 1568 m. For the interval during which the train is slowing down, v =.4 m/s, a = 3.5 m/s and v = and v = v + a ( ) gives = v v = (.4 m/s) = 7 m. a ( 3.5 m/s ) The total distance traveled is 157 m m + 7 m = 18 m. EVALUATE: The acceleration is not constant for the entire motion, but it does consist of constant acceleration segments, and we can use constant acceleration equations for each segment..61. IDENTIFY: When the graph of v versus t is a straight line the acceleration is constant, so this motion consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. Since v is alwas positive the motion is alwas in the + direction and the total distance moved equals the magnitude of the displacement. The acceleration a is the slope of the v versus t graph. SET UP: For the t = to t = 1. s 1. s segment, v = 1. m/s and v =. segment, v = 4. m/s and v = 1. m/s. For the t = 1. s to v EXECUTE: (a) For t = to t = 1. s, = + v t = 4. m/s + 1. m/s (1. s) = 8. m. 1. m/s + For t = 1. s to t = 1. s, = (. s) = 1. m. The total distance traveled is 9. m. (b) = 8. m + 1. m = 9. m =.

25 Copright 16 Pearson Education, Inc. 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.

26 - Chapter (c) For t = to 1. s, a = 1. m/s 4. m/s =.8 m/s. For t = 1. s to 1. s, 1. s a = 1. m/s = 6. m/s. The graph of a versus t is given in Figure.61.. s EVALUATE: When v and a are both positive, the speed increases. When v is positive and a is negative, the speed decreases. Figure IDENTIFY: Appl = v t + 1 a t to the motion of each train. A collision means the front of the passenger train is at the same location as the caboose of the freight train at some common time. SET UP: Let P be the passenger train and F be the freight train. For the front of the passenger train = and for the caboose of the freight train = m. For the freight train v F = 15. m/s and a F =. For the passenger train v P = 5. m/s and a P =.1 m/s. EXECUTE: (a) = v t + 1 a t for each object gives P = v P t + 1 a Pt and F = m + v Ft. Setting = F gives v t + 1 a t = m + v t. (. 5 m/s )t (1. m/s)t + m =. The quadratic P P P F 1 formula gives t = ± (1. ) 4(. 5)() s = (1 ± 77.5) s. The collision occurs at t = 1 s s =.5 s. The equations that specif a collision have a phsical solution (real, positive t), so a collision does occur. (b) P = (5. m/s)(. 5 s) + 1 (. 1 m/s )(. 5 s) = 537 m. The passenger train moves 537 m before the collision. The freight train moves (15. m/s)(. 5 s) = 337 m. (c) The graphs of F and P versus t are sketched in Figure.6. EVALUATE: The second root for the equation for t, t = s is the time the trains would meet again if the were on parallel tracks and continued their motion after the first meeting. Figure IDENTIFY and SET UP: Appl constant acceleration kinematics equations. Find the velocit at the start of the second 5. s; this is the velocit at the end of the first 5. s. Then find for the first 5. s. Copright 16 Pearson Education, Inc. 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.

27 Motion Along a Straight Line -3 EXECUTE: For the first 5. s of the motion, v =, t = 5. s. v = v + a t gives v = a (5. s). This is the initial speed for the second 5. s of the motion. For the second 5. s: v = a (5. s), t = 5. s, = m. = v t + 1 a t gives m = (5 s )a + (1.5 s )a so a = m/s. Use this a and consider the first 5. s of the motion: = v t + 1 a t = + 1 (5.333 m/s )(5. s) = 67 m. EVALUATE: The ball is speeding up so it travels farther in the second 5. s interval than in the first..64. IDENTIFY: The insect has constant speed 15 m/s during the time it takes the cars to come together. SET UP: Each car has moved 1 m when the hit. EXECUTE: The time until the cars hit is 1 m = 1 s. During this time the grasshopper travels a 1 m/s distance of (15 m/s)(1 s) = 15 m. EVALUATE: The grasshopper ends up 1 m from where it started, so the magnitude of his final displacement is 1 m. This is less than the total distance he travels since he spends part of the time moving in the opposite direction..65. IDENTIFY: Appl constant acceleration equations to each object. Take the origin of coordinates to be at the initial position of the truck, as shown in Figure.65a. Let d be the distance that the car initiall is behind the truck, so (car) = d and (truck) =. Let T be the time it takes the car to catch the truck. Thus at time T the truck has undergone a displacement = 6. m, so is at = + 6. m = 6. m. The car has caught the truck so at time T is also at = 6. m. Figure.65a (a) SET UP: Use the motion of the truck to calculate T: = 6. m, v = (starts from rest), a =.1 m/s, t = T = v t + 1 a t ( ) Since v =, this gives t = a EXECUTE: T = (6. m) = 7.56 s.1 m/s (b) SET UP: Use the motion of the car to calculate d: = 6. m + d, v =, a = 3.4 m/s, t = 7.56 s = v t + 1 a t EXECUTE: d + 6. m = 1 (3.4 m/s )(7.56 s) d = m 6. m = 37. m. (c) car: v = v + a t = + (3. 4 m/s )(7.56 s) = 5.7 m/s truck: v = v + a t = + (. 1 m/s )(7.56 s) = 15.9 m/s Copright 16 Pearson Education, Inc. All rights reserved. This material is protected under all copright laws as the currentl eist.