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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 positiontime 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 positiontime 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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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