Physics Solution Manual for 1100 and 2101

0 m 980 ms 2 175 s 1 note that the stone is falling

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Unformatted text preview: TION To find the initial velocity v0,2 of the second stone, we employ Equation 2.8, y = v0,2t2 + 1 at2 2 . Solving this equation for v0,2 yields 2 v0,2 = y − 1 at2 2 2 t2 The time t1 for the first stone to strike the ground can be obtained from Equation 2.8, y = v0,1t1 + 1 2 at12 . Noting that v0,1 = 0 m/s since the stone is dropped from rest and solving this equation for t1, we have t1 = 2y = a 2 ( −15.0 m ) −9.80 m/s 2 = 1.75 s (1) Note that the stone is falling down, so its displacement is negative (y = − 15.0 m). Also, its 2 acceleration a is that due to gravity, so a = −9.80 m/s . The time t3.20 for the first stone to fall 3.20 m can also be obtained from Equation 1: t3.20 = 2y = a 2 ( −3.20 m ) −9.80 m/s 2 The time t2 that the second stone is in the air is = 0.808 s Chapter 2 Problems 81 t2 = t1 − t3.20 = 1.75 s − 0.808 s = 0.94 s The initial velocity of the second stone is v0,2 = y − 1 at2 2 2 ( −15.0 m ) − 1 ( −9.80 m/s2 ) ( 0.94 s )2 2 = = −11 m/s t2 0.94 s ______________________________________________________________________________ 64. REASONING AND SOLUTION We measure the positions of the balloon and the pellet relative to the ground and assume up to be positive. The balloon has no acceleration, since it travels at a constant velocity vB, so its displacement in time t is vBt. Its position above the ground, therefore, is yB = H 0 + vBt where H0 = 12 m. The pellet moves under the influence of gravity (a = –9.80 m/s2), so its position above the ground is given by Equation 2.8 as yP = v0t + 1 at 2 2 But yP = yB at time t, so that v0t + 1 at 2 = H 0 + vBt 2 Rearranging this result and suppressing the units gives 1 at 2 2 + ( v0 – vB ) t – H 0 = 1 2 ( –9.80 ) t 2 + ( 30.0 – 7.0 ) t –12.0 = 0 4.90t 2 – 23.0t + 12.0 = 0 t= 23.0 ± 23.02 – 4 ( 4.90 )(12.0 ) 2 ( 4.90 ) = 4.09 s or 0.602 s Substituting each of these values in the expression for yB gives yB = 12.0 m + ( 7.0 m/s ) ( 4.09 s ) = 41 m yB = 12.0 m + ( 7.0 m/s )( 0.602 s ) = 16 m ______________________________________________________________________________ 82 KINEMATICS IN ONE DIMENSION 65. REASONING AND SOLUTION The average acceleration for each segment is the slope of that segment. 40 m/s − 0 m/s aA = = 1.9 m/s 2 21 s − 0 s aB = 40 m/s − 40 m/s = 0 m/s 2 48 s − 21 s 80 m/s − 40 m/s = 3.3 m/s 2 60 s − 48 s ______________________________________________________________________________ aC = 66. REASONING On a position-versus-time graph, the velocity is the slope. Since the object’s velocity is constant and it moves in the +x direction, the graph will be a straight line with a positive slope, beginning at x = −16 m when t = 0 s. At t = 18 s, its position should be x = −16 m + 48 m = +32 m. Once the graph is constructed, the object’s velocity is found by ∆x . calculating the slope of the graph: v = ∆t SOLUTION The position-versus-time graph for the motion is as follows: 40 Position x (m) 32 24 16 8.0 0.0 −8.0 −16 −24 0 3.0 6.0 9.0 12 15 18 Time t (s) The object’s displacement is +48 m, and the elapsed time is 18 s, so its velocity is v= 67. ∆ x +4 8 m = = +2.7 m/s ∆t 18 s SSM REASONING The slope of a straight-line segment in a position-versus-time graph is the average velocity. The algebraic sign of the average velocity, therefore, corresponds to the sign of the slope. Chapter 2 Problems 83 SOLUTION a. The slope, and hence the average velocity, is positive for segments A and C, negative for segment B, and zero for segment D. b. In the given position-versus-time graph, we find the slopes of the four straight-line segments to be 1.25 km − 0 km vA = = +6.3 km/h 0.20 h − 0 h vB = 0.50 km − 1.25 km = −3.8 km/h 0.40 h − 0.20 h vC = 0.75 km − 0.50 km = +0.63 km/h 0.80 h − 0.40 h 0.75 km − 0.75 km = 0 km/h 1.00 h − 0.80 h ______________________________________________________________________________ vD = 68. REASONING segment. The average velocity for each segment is the slope of the line for that SOLUTION Taking the direction of motion as positive, we have from the graph for segments A, B, and C, 10.0 km – 40.0 km vA = = –2.0 × 101 km/h 1.5 h – 0.0 h vB = 20.0 km – 10.0 km = 1.0 × 101 km/h 2.5 h – 1.5 h 40.0 km – 20.0 km = 40 km/h 3.0 h – 2.5 h ______________________________________________________________________________ vC = 69. REASONING The slope of the position-time graph is the velocity of the bus. Each of the three segments of the graph is a straight line, so the bus has a different constant velocity for each part of the trip: vA, vB, and vC . The slope of each segment may be calculated from ∆x Equation 2.2 v = , where ∆x is the difference between the final and initial positions of ∆t the bus and ∆t is the elapsed time during each segment. The average acceleration of the bus v − v0 is the change in its velocity divided by the elapsed time, as in Equation 2.4 a = . ∆t The trip lasts from t = 0 h (the initial instant on the graph) to t = 3.5 h (the final insta...
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