Physics Solution Manual for 1100 and 2101

If the full displacements for each segment are chosen

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Unformatted text preview: nt on the graph), so the total elapsed time is ∆t = 3.5 h. The initial velocity of the bus is its 84 KINEMATICS IN ONE DIMENSION velocity at t = 0, which is its constant velocity for segment A: v0 = vA. Similarly, the velocity of the bus at the last instant of segment C is its final velocity for the trip: v = vC. ∆x SOLUTION In using Equation 2.2 v = to calculate the slopes of segments A and C, ∆t any displacement ∆x within a segment may be chosen, so long as the corresponding elapsed time ∆t is used in the calculation. If the full displacements for each segment are chosen, then ∆x A 24 km − 0 km = = 24 km/h ∆t A 1.0 h − 0 h vA = vC = ∆xC ∆tC = 27 km − 33 km = −5 km/h 3.5 h − 2.2 h Apply these results to Equation 2.4: a= 70. REASONING (v 2 ) v − v0 ∆t = ( −5 km/h ) − ( 24 km/h ) = 3.5 h −8.3 km/h 2 The car’s initial velocity v0 may be calculated from Equation 2.9 2 = v0 + 2ax , since we know the final velocity v = +4.0 m/s, the displacement x = +120 m, and the acceleration a = −3.2 m/s2. The acceleration is negative for the following reason: the car is slowing down, so the acceleration points opposite to the velocity, which is positive. To construct the graph, we need to know the elapsed time t, which may be found from the initial and final velocities and Equation 2.4 ( v = v0 + at ) . Because the car’s acceleration is constant and negative, it should be a straight line with a negative slope. SOLUTION ( 2 2 ) a. Solving Equation 2.9 v = v0 + 2ax for the initial velocity gives 2 v0 = v 2 − 2ax v0 = + v 2 − 2ax = + ( 4.0 m/s )2 − 2 ( −3.2 m/s )(120 m ) = +28 m/s The positive root is appropriate, since the car is traveling in the +x direction. Chapter 2 Problems 85 b. We know that at t = 0 s, the car’s velocity is +28 m/s. Equation 2.4 ( v = v0 + at ) can be solved to give the time necessary to slow down to +4.0 m/s: t= v − v0 a = 4.0 m/s − 28 m/s −3.2 m/s 2 = 7.5 s Therefore, the graph should run from t = 0 s to t = 7.5 s: 32 Velocity v (m/s) 28 24 20 16 12 8.0 4.0 0 0 1.5 3.0 4.5 6.0 7.5 Time t (s) 71. SSM REASONING The two runners start one hundred meters apart and run toward each other. Each runs ten meters during the first second and, during each second thereafter, each runner runs ninety percent of the distance he ran in the previous second. While the velocity of each runner changes from second to second, it remains constant during any one second. SOLUTION The following table shows the distance covered during each second for one of the runners, and the position at the end of each second (assuming that he begins at the origin) for the first eight seconds. Time t (s) Distance covered (m) Position x (m) 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 10.00 9.00 8.10 7.29 6.56 5.90 5.31 4.78 0.00 10.00 19.00 27.10 34.39 40.95 46.85 52.16 56.94 86 KINEMATICS IN ONE DIMENSION The following graph is the position-time graph constructed from the data in the table above. 60 50 40 30 20 10 0 0 2 4 6 8 10 Time t (s) a. Since the two runners are running toward each other in exactly the same way, they will meet halfway between their respective starting points. That is, they will meet at x = 50.0 m. According to the graph, therefore, this position corresponds to a time of 6.6 s . b. Since the runners collide during the seventh second, the speed at the instant of collision can be found by taking the slope of the position-time graph for the seventh second. The speed of either runner in the interval from t = 6.00 s to t = 7.00 s is v= ∆x 52.16 m – 46.85 m = = 5.3 m/s ∆t 7.00 s – 6.00 s Therefore, at the moment of collision, the speed of either runner is 5.3 m/s . ______________________________________________________________________________ 72. REASONING AND SOLUTION Since v = v0 + at, the acceleration is given by a = ( v – v0 ) / t . Since the direction of travel is in the negative direction throughout the problem, all velocities will be negative. a. a= ( − 29.0 m/s) − ( − 27.0 m/s) = −0.40 m/s 2 5.0 s Since the acceleration is negative, it is in the same direction as the velocity and the car is speeding up. Chapter 2 Problems b. a= 87 ( − 23.0 m/s) − ( − 27.0 m/s) = +0.80 m/s 2 5.0 s Since the acceleration is positive, it is in the opposite direction to the velocity and the car is slowing down or decelerating. ______________________________________________________________________________ 73. SSM REASONING AND SOLUTION When air resistance is neglected, free fall conditions are applicable. The final speed can be found from Equation 2.9; 2 v 2 = v0 + 2ay where v0 is zero since the stunt man falls from rest. If the origin is chosen at the top of the hotel and the upward direction is positive, then the displacement is y = –99.4 m. Solving for v, we have v = − 2ay = – 2(–9.80 m/s 2 )(–99.4 m) = – 44.1 m/s The speed at impact is the magnitude of this result or 44.1 m/s . ______________________________________________________________________________ 74. REASONING AND...
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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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