Chapter_4Soln

# Chapter_4Soln - Solutions to Assigned Problems Unit 2...

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1 1 Solutions to Assigned Problems, Unit 2, Chapter 4 MOTION WITH A CHANGING VELOCITY 1. (a) Strategy Relate the acceleration, speed, and distance using Eq. (4-5). Solution 22 2 2 fi 0( 3 . 2 m s ) 2 , so 0.85 m s . 2 2(6.0 m 0) xx x x vv a x a x −= Δ = = = Δ− The magnitude of the average acceleration is 2 0.85 m s . (b) Strategy Use Newton’s second law. Draw a free-body diagram. S o l u t i o n Find k . μ kk k k or . x x ma F f ma f N ma N μμ ∑= = = = 0, so . y F Nm g N m g = = Thus, we have 2 k 2 0.85 m s 0.087 . 9.80 m s x ma ma a g g =− = f N m g y x v k 5. (a) Strategy Use Eqs. (3-3) and (4-2) since the acceleration is constant. Solution Find the distance traveled. av, 27.3 m s 17.4 m s (10.0 s) 224 m . x xv t t + + Δ= = (b) Strategy Use the definition of average acceleration. Solution Find the magnitude of the acceleration. 2 27 .3 ms 17 .4 ms 0.99 m s 10.0 s v a t == = Δ 9. Strategy Use Eq. (4-5) to find the distance the car requires to stop after slamming on the brakes. Then, use Eqs. (3-3) and (4-2) to relate the speeds, distances, time, and acceleration. Solution Find the distance the car requires to stop. 2 fc ic c 2 2 7 . 0 m s ) 52.1 m . 2 2( 7.00 m s ) x a = = Since the acceleration of the car is constant, the average speed of the car as it attempts to stop is c,av fc ic ( )2 (0 27 .0 )2 13 .5 . v = += + = Thus, the time required for the car to stop is cc , a v (52.1 m) (13.5 m s) 3.86 s. tx v Δ=Δ = = The distance the tractor travels in this time is tt (10.0 m s)(3.86 s) 38.6 m. t Δ=Δ= = Now, 38.6 m 15.0 m 53.6 m, + = which is 53.6 m 52.1 m 1.5 m beyond the stopping point of the car. Therefore, you won’t hit the tractor.

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Chapter 4: Motion with a Changing Velocity College Physics 2 13. (a) Strategy The graph will be a line with a slope of 2 1.20 m s . Solution 0 when 0. x vt == The graph is shown. (b) Strategy Use Eq. (4-4). Solution Find the distance the train traveled. 22 2 i 11 1 () ( 0 ) () 2 1 (1.20 m s )(12.0 s) 86.4 m 2 xx x x x vt a t t a t a t Δ = Δ+ Δ = Δ+ Δ = Δ v x (m/s) t (s) 2 4 6 8 10 12 14 0 2 4 6 8 10 12 16 0 (c) Strategy Use Eq. (4-1). Solution Find the final speed of the train. 2 fi f f 0 , so (1.20 m s )(12.0 s) 14.4 m s . x x vvv a tv a t −=− = Δ = Δ = = 17. Strategy Refer to the figure. Analyze graphically and algebraically. Solution Graphical analysis: Find the slope of the graph. 2 av, 40 m s 20 m s 5.0 m s 9.0 s 5.0 s x a Algebraic solution: 2 a v , a v , 40 m s 20 m s , so 5.0 m s .
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Chapter_4Soln - Solutions to Assigned Problems Unit 2...

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