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Unformatted text preview: hr = 1 day and 3600 s = 1 hr, the acceleration of the spacecraft (in m/s2) is
+9.0 m / s
= +1.04 × 10−4 m / s 2 24 hr 3600 s (1 day ) 1 day 1 hr ______________________________________________________________________________
a= 18. REASONING ∆v
=
t We can use the definition of average acceleration a = ( v − v 0 ) / ( t − t0 ) (Equation 2.4) to find the sprinter’s final velocity v at the end of the acceleration phase,
because her initial velocity ( v 0 = 0 m/s , since she starts from rest), her average acceleration a , and the time interval t − t0 are known.
SOLUTION
a. Since the sprinter has a constant acceleration, it is also equal to her average acceleration,
so a = +2.3 m/s 2 Her velocity at the end of the 1.2s period is v = v0 + a ( t − t0 ) = ( 0 m/s ) + ( +2.3 m/s 2 ) (1.2 s ) = +2.8 m/s
b. Since her acceleration is zero during the remainder of the race, her velocity remains
constant at +2.8 m/s .
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19. REASONING When the velocity and acceleration vectors are in the same direction, the
speed of the object increases in time. When the velocity and acceleration vectors are in
opposite directions, the speed of the object decreases in time. (a) The initial velocity and
acceleration are in the same direction, so the speed is increasing. (b) The initial velocity and
acceleration are in opposite directions, so the speed is decreasing. (c) The initial velocity
and acceleration are in opposite directions, so the speed is decreasing. (d) The initial
velocity and acceleration are in the same direction, so the speed is increasing. SOLUTION The final velocity v is related to the initial velocity v0, the acceleration a, and
the elapsed time t through Equation 2.4 (v = v0 + a t). The final velocities and speeds for the
four moving objects are: 54 KINEMATICS IN ONE DIMENSION a. v = 12 m/s + (3.0 m/s2)(2.0 s) = 18 m/s. The final speed is 18 m/s .
b. v = 12 m/s + (−3.0 m/s2)(2.0 s) = 6.0 m/s. The final speed is 6.0 m/s .
c. v = −12 m/s + (3.0 m/s2)(2.0 s) = −6.0 m/s. The final speed is 6.0 m/s .
d. v = −12 m/s + (−3.0 m/s2)(2.0 s) = −18 m/s. The final speed is 18 m/s .
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20. REASONING AND SOLUTION
from Equation 2.4 (v = v0 + at) as
v − v0 The magnitude of the car's acceleration can be found 26.8 m/s – 0 m/s
= 8.18 m/s 2
t
3.275 s
______________________________________________________________________________ a= 21. = SSM REASONING AND SOLUTION The velocity of the automobile for each stage is
given by Equation 2.4: v = v0 + at . Therefore,
v1 = v0 + a1t = 0 m/s + a1t and v2 = v1 + a2t Since the magnitude of the car's velocity at the end of stage 2 is 2.5 times greater than it is at
the end of stage 1, v2 = 2.5v1 . Thus, rearranging the result for v2, we find
v2 – v1 2.5v1 – v1 1.5v1 1.5( a1t )
=
=
=
= 1.5a1 = 1.5(3.0 m/s 2 ) = 4.5 m/s 2
t
t
t
t
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a2 = 22. REASONING According to Equation 2.4, the average acceleration of the car for the first
twelve seconds after the engine cuts out is
a1 = v1f − v10 (1) ∆t1 and the average acceleration of the car during the next six seconds is
a2 = v2f − v20
∆t2 = v2f − v1f
∆t2 (2) The velocity v1f of the car at the end of the initial twelvesecond interval can be found by
solving Equations (1) and (2) simultaneously. Chapter 2 Problems 55 SOLUTION Dividing Equation (1) by Equation (2), we have
a1
a2 = (v1f − v10 ) / ∆t1 (v2f − v1f ) / ∆t2 = (v1f − v10 )∆t2 (v2f − v1f )∆t1 Solving for v1f , we obtain
v1f = v1f = a1∆t1v2f + a2 ∆t2v10
a1∆t1 + a2 ∆t2 = (a1 / a2 )∆t1v2f + ∆t2v10
(a1 / a2 )∆t1 + ∆t2 1.50(12.0 s)(+28.0 m/s) + (6.0 s)( + 36.0 m/s)
= +30.0 m/s
1.50(12.0 s) + 6.0 s 23. REASONING AND SOLUTION Both motorcycles have the same velocity v at the end of
the four second interval. Now
v = v0A + aAt
for motorcycle A and
v = v0B + aBt
for motorcycle B. Subtraction of these equations and rearrangement gives
v0A – v0B = (4.0 m/s2 – 2.0 m/s2)(4 s) = +8.0 m/s The positive result indicates that motorcycle A was initially traveling faster.
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24. REASONING AND SOLUTION
a = v / t , so The average acceleration of the basketball player is 2 6.0 m/s x = 1 at 2 = 1 (1.5 s ) = 4.5 m
2
2
1.5 s ______________________________________________________________________________ 25. SSM REASONING AND SOLUTION ( found by solving Equation 2.9 v 2 2
= v0 The average acceleration of the plane can be ) + 2ax for a. Taking the direction of motion as positive, we have
a= 2
v 2 − v0 2x = (+6.1 m/s) 2 − (+69 m/s)2
= – 3.1 m/s 2
2(+750 m) 56 KINEMATICS IN ONE DIMENSION The minus sign indicates that the direction of the acceleration is opposite to the direction of
motion, and the plane is slowing down.
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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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