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26. REASONING The average acceleration is defined by Equation 2.4 as the change in
velocity divided by the elapsed time. We can find the elapsed time from this relation
because the acceleration and the change in velocity are given.
SOLUTION
a. The time ∆t that it takes for the VW Beetle to change its velocity by an amount
∆v = v – v0 is (and noting that 0.4470 m/s = 1 mi/h) ∆t = v − v0 = a 0.4470 m / s −0 m/s 1 mi / h ( 60.0 mi / h ) 2.35 m / s 2 = 11.4 s b. From Equation 2.4, the acceleration (in m/s2) of the dragster is 0.4470 m / s −0 m/s
v − v0 1 mi / h a=
=
= 44.7 m / s 2
t − t0
0.600 s − 0 s
______________________________________________________________________________ ( 60.0 mi / h ) 27. REASONING We know the initial and final velocities of the blood, as well as its ( ) 2
displacement. Therefore, Equation 2.9 v 2 = v0 + 2ax can be used to find the acceleration of the blood. The time it takes for the blood to reach it final velocity can be found by using x
Equation 2.7 t = 1
.
v + v) 2( 0 SOLUTION
a. The acceleration of the blood is a= 2
v 2 − v0 2x ( 26 cm / s )2 − ( 0 cm / s )2
=
2 ( 2.0 cm ) = 1.7 × 102 cm / s 2 b. The time it takes for the blood, starting from 0 cm/s, to reach a final velocity of +26 cm/s
is
x
2.0 cm
t= 1
=1
= 0.15 s
( v0 + v ) 2 ( 0 cm / s + 26 cm / s )
2
______________________________________________________________________________ Chapter 2 Problems 57 28. REASONING AND SOLUTION
a. From Equation 2.4, the definition of average acceleration, the magnitude of the average
acceleration of the skier is
a= v − v0
t − t0 = 8.0 m/s – 0 m/s
= 1.6 m/s 2
5.0 s b. With x representing the displacement traveled along the slope, Equation 2.7 gives:
x = 1 (v0 + v )t = 1 (8.0 m/s + 0 m/s)(5.0 s) = 2.0 × 101 m
2
2 ______________________________________________________________________________
29. SSM WWW REASONING AND SOLUTION
a. The magnitude of the acceleration can be found from Equation 2.4 (v = v0 + at) as
a= v − v0
t = 3.0 m/s – 0 m/s
= 1.5 m/s 2
2.0 s b. Similarly the magnitude of the acceleration of the car is
a= v − v0
t = 41.0 m/s – 38.0 m/s
= 1.5 m/s 2
2.0 s c. Assuming that the acceleration is constant, the displacement covered by the car can be
found from Equation 2.9 (v2 = v02 + 2ax): x= 2
v 2 − v0 2a = (41.0 m/s)2 − (38.0 m/s)2
= 79 m
2(1.5 m/s 2 ) Similarly, the displacement traveled by the jogger is x= 2
v 2 − v0 2a = (3.0 m/s) 2 − (0 m/s) 2
= 3.0 m
2(1.5 m/s 2 ) Therefore, the car travels 79 m – 3.0 m = 76 m further than the jogger.
______________________________________________________________________________ 58 KINEMATICS IN ONE DIMENSION 30. REASONING The cheetah and its prey run the same distance. The prey runs at a constant
velocity, so that its distance is the magnitude of its displacement, which is given by
Equation 2.2 as the product of velocity and time. The distance for the cheetah can be
expressed using Equation 2.8, since the cheetah’s initial velocity (zero, since it starts from
rest) and the time are given, and we wish to determine the acceleration. The two expressions
for the distance can be equated and solved for the acceleration.
SOLUTION We begin by using Equation 2.2 and assuming that the initial position of the
prey is x0 = 0 m. The distance run by the prey is
∆x = x − x0 = x = vPreyt The distance run by the cheetah is given by Equation 2.8 as
x = v0, Cheetah t + 1 aCheetah t 2
2 Equating the two expressions for x and using the fact that v0, Cheetah = 0 m/s, we find that vPreyt = 1 aCheetah t 2
2
Solving for the acceleration gives
aCheetah = 31. 2vPrey
t = 2 ( +9.0 m/s )
3.0 s = +6.0 m/s 2 SSM WWW REASONING Since the belt is moving with constant velocity, the
displacement (x0 = 0 m) covered by the belt in a time tbelt is giving by Equation 2.2 (with x0
assumed to be zero) as
x = vbelt tbelt (1 ) Since Clifford moves with constant acceleration, the displacement covered by Clifford in a
time tCliff is, from Equation 2.8,
2
2
x = v0tCliff + 1 atCliff = 1 atCliff
2
2 (2 ) The speed vbelt with which the belt of the ramp is moving can be found by eliminating x
between Equations (1) and (2). Chapter 2 Problems 59 SOLUTION Equating the right hand sides of Equations (1) and (2), and noting that
tCliff = 1 tbelt , we have
4 vbelt tbelt = 1 a
2
vbelt = 1 at
32 belt ( 1 tbelt )
4 2 1
= 32 (0.37 m/s 2 )(64 s) = 0.74 m/s ______________________________________________________________________________
32. REASONING At time t both rockets return to their starting points and have a displacement
of zero. This occurs, because each rocket is decelerating during the first half of its journey.
However, rocket A has a smaller initial velocity than rocket B. Therefore, in order for
rocket B to decelerate and return to its point of origin in the same time as rocket A, rocket B
must have a deceleration with a greater magnitude than that for rocket A. Since we know
that the displacement of each rocket is zero at time t, since both initial velocities are given,
and since we seek information about the acceleration, we begin our solution with Equation
2.8, for it contains just these variables.
SOLUTION Applying Equation 2.8 to each rocket gives xA = v0A t + 1 aA t 2
2 xB = v0Bt + 1 aBt 2
2 0 = v0A t + 1 aA t 2
2 0 = v0Bt + 1 aBt 2
2...
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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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