Physics Solution Manual for 1100 and 2101

# 8003 180 ms 0800 s 180 ms t 1 500 ms2

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Unformatted text preview: ons, the part due to the constant-velocity motion during the reaction time and the part due to the accelerated motion. Using Equation 2.2 for the contribution from the constant-velocity motion and Equation 2.9 for the contribution from the accelerated motion, we obtain xPolice car = v0, Police car ( 0.800 s ) + v0, Police car t + 1 at 2 23 1444 2444 3 144 2444 4 Constant velocity motion, Equation 2.2 Accelerated motion, Equation 2.8 ( ) = (18.0 m/s )( 0.800 s ) + (18.0 m/s ) t + 1 5.00 m/s 2 t 2 2 Setting the two displacements equal we obtain s) ( 42.0 m/s )( t + 0.8003 = (18.0 m/s )( 0.800 s ) + (18.0 m/s ) t + 1 ( 5.00 m/s2 ) t 2 2 1444 24444 4 Displacement of speeder 144444444424444444444 4 3 Displacement of police car Rearranging and combining terms gives this result in the standard form of a quadratic equation: 94 KINEMATICS IN ONE DIMENSION ( 2.50 m/s2 ) t 2 − ( 24.0 m/s ) t − 19.2 m = 0 Solving for t shows that t= − ( −24.0 m/s ) ± ( −24.0 m/s )2 − 4 ( 2.50 m/s2 ) ( −19.2 m ) ( 2 2.50 m/s 2 ) = 10.3 s We have ignored the negative root, because it leads to a negative value for the time, which is unphysical. The total time for the police car to catch up, including the reaction time, is 0.800 s + 10.3 s = 11.1 s 86. REASONING AND SOLUTION The balls pass at a time t when both are at a position y above the ground. Applying Equation 2.8 to the ball that is dropped from rest, we have y = 24 m + v01t + 1 at 2 = 24 m + ( 0 m/s ) t + 1 at 2 2 2 (1) Note that we have taken into account the fact that y = 24 m when t = 0 s in Equation (1). For the second ball that is thrown straight upward, y = v02t + 1 at 2 2 (2) Equating Equations (1) and (2) for y yields 24 m + 1 at 2 = v02t + 1 at 2 2 2 or 24 m = v02t Thus, the two balls pass at a time t, where t= 24 m v02 The initial speed v02 of the second ball is exactly the same as that with which the first ball hits the ground. To find the speed with which the first ball hits the ground, we take upward ( ) 2 as the positive direction and use Equation 2.9 v 2 = v0 + 2ay . Since the first ball is dropped from rest, we find that v02 = v = 2ay = 2(–9.80 m/s 2 )(–24 m) = 21.7 m/s Chapter 2 Problems 95 Thus, the balls pass after a time t= 24 m = 1.11 s 21.7 m/s At a time t = 1.11 s, the position of the first ball according to Equation (1) is 1 2 y = 24 m + (–9.80 m/s 2 )(1.11 s) 2 = 24 m – 6.0 m which is 6.0 m below the top of the cliff . ______________________________________________________________________________ 87. SSM REASONING Since the car is moving with a constant velocity, the displacement of the car in a time t can be found from Equation 2.8 with a = 0 m/s2 and v0 equal to the velocity of the car: xcar = vcar t . Since the train starts from rest with a constant acceleration, the displacement of the train in a time t is given by Equation 2.8 with v0 = 0 m/s: xtrain = 1 atrain t 2 2 At a time t1 , when the car just reaches the front of the train, xcar = Ltrain + xtrain , where Ltrain is the length of the train. Thus, at time t1 , 2 vcar t1 = Ltrain + 1 atrain t1 2 (1) At a time t2, when the car is again at the rear of the train, xcar = xtrain . Thus, at time t2 2 vcar t2 = 1 atrain t2 2 (2) Equations (1) and (2) can be solved simultaneously for the speed of the car vcar and the acceleration of the train atrain . SOLUTION a. Solving Equation (2) for atrain we have atrain = 2vcar t2 Substituting this expression for atrain into Equation (1) and solving for vcar , we have (3) 96 KINEMATICS IN ONE DIMENSION vcar = Ltrain t t1 1 − 1 t 2 = 92 m = 13 m/s 14 s (14 s) 1 − 28 s b. Direct substitution into Equation (3) gives the acceleration of the train: atrain = 2vcar = 2 (13 m/s) = 0.93 m/s 2 28 s t2 ______________________________________________________________________________ 88. REASONING AND SOLUTION During the first phase of the acceleration, a1 = v t1 During the second phase of the acceleration, v = (3.4 m/s) – (1.1 m/s2)(1.2 s) = 2.1 m/s Then 2.1 m/s = 1.4 m/s 2 1.5 s ______________________________________________________________________________ a1 = CHAPTER 3 KINEMATICS IN TWO DIMENSIONS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (a) The horizontal component vx of the projectile’s velocity remains constant throughout the motion, since the acceleration ax in the horizontal direction is zero (ax = 0 m/s2). The vertical component vy, however, changes as the projectile moves. This component is greatest at point 1, decreases to zero at point 2 at the top of the trajectory, and then increases to a magnitude less than that at point 1 as the projectile approaches point 3. 2. (b) The minimum speed of the projectile occurs when it is at the top of its trajectory. At this point the vertical component of its velocity is zero (vy = 0 m/s). Since there is no 2 acceleration in the x direction (ax = 0 m/s ), the x component of the projectile’s velocity remains constant at vx = +30 m/s throughout the motion. Thus, the minimum speed is 30 m/s. 3. (c) The acceleration du...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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