Unformatted text preview: positive direction). Sir Alfred’s acceleration is negative ( aA = −0.200 m/s2 ) since he starts from rest and moves to the left (the negative direction). The displacement of Sir George is, then,
xG = d
88.0 m
=
= 52.8 m
aA
( −0.200 m/s2 )
1−
1−
aG
( +0.300 m/s2 ) ____________________________________________________________________________________________ 36. REASONING At a constant velocity the time required for the first car to travel to the next
exit is given by Equation 2.2 as the magnitude of the displacement (2.5 × 103 m) divided by
the magnitude of the velocity. This is also the travel time for the second car to reach the next
exit. The acceleration for the second car can be determined from Equation 2.8, since the
initial velocity, the displacement, and the time are known. This equation applies, because
the acceleration is constant.
SOLUTION According to Equation 2.2, with the assumption that the initial time is t0 = 0 s,
the time for the first car to reach the next exit at a constant velocity is
∆t = t − t 0 = t = ∆x 2.5 × 103 m
=
= 76 s
v
33 m/s Remembering that the initial velocity v0 of the second car is zero, we can solve Equation 2.8 ( x = v0t + 12 at 2 = 1 at 2 ) for the acceleration to show that
2
a= 2x
t 2 = 2(2.5 × 103 m)
(76 s) Since the second car’s speed
in the same direction as the velocity . is 2 = 0.87 m/s 2 increasing, this acceleration must be 37. REASONING AND SOLUTION The speed of the car at the end of the first (402 m) phase
can be obtained as follows:
v12 = v02 + 2a1x1
v1 = 2 (17.0 m/s 2 ) ( 402 m ) 64 KINEMATICS IN ONE DIMENSION The speed after the second phase (3.50 × 102 m) can be obtained in a similar fashion.
v22 = v022 + 2a2x2
2
v2 = v1 + 2 ( −6.10 m/s2 )( 3.50 ×102 m ) v2 = 96.9 m/s
______________________________________________________________________________
38. REASONING AND SOLUTION
a. The velocity at the end of the first (7.00 s) period is
v1 = v0 + a1 = (2.01 m/s2)(7.00 s)
At the end of the second period the velocity is
v2 = v1 + a2t2 = v1 + (0.518 m/s2)(6.00 s)
And the velocity at the end of the third (8.00 s) period is
v3 = v2 + a3t3 = v2 + (–1.49 m/s2)(8.00 s) = +5.26 m/s
b. The displacement for the first time period is found from
x1 = v0t1 +
x1 = (0 m/s)(7.00 s) + 1 (2.01
2 1a t 2
2 11 m/s2)(7.00 s)2 = +49.2 m Similarly, x2 = +93.7 m and x3 = +89.7 m, so the total displacement of the boat is
x = x1 + x2 + x3 = +233 m ______________________________________________________________________________
39. REASONING Because the car is traveling in the +x direction and decelerating, its
acceleration is negative: a = −2.70 m/s2. The final velocity for the interval is given
(v = +4.50 m/s), as well as the elapsed time (t = 3.00 s). Both the car’s displacement x and
its initial velocity v0 at the instant braking begins are unknown. Chapter 2 Problems 65 Compare the list of known kinematic quantities (v, a, t) to the equations of kinematics for
constant acceleration: v = v0 + at (Equation 2.4), x = 1 ( v0 + v ) t (Equation 2.7),
2
2
x = v0t + 1 at 2 (Equation 2.8), and v 2 = v0 + 2 ax (Equation 2.9). None of these four
2 equations contains all three known quantities and the desired displacement x, and each of
them contains the initial velocity v0. Since the initial velocity is neither known nor
requested, we can combine two kinematic equations to eliminate it, leaving an equation in
which x is the only unknown quantity.
SOLUTION For the first step, solve Equation 2.4 ( v = v0 + at ) for v0:
v0 = v − at (1) ( ) Substituting the expression for v0 in Equation (1) into Equation 2.8 x = v0t + 1 at 2 yields
2
an expression for the car’s displacement solely in terms of the known quantities v, a, and t:
x = ( v − at ) t + 1 at 2 = vt − at 2 + 1 at 2
2
2
x = vt − 1 at 2
2 (2) Substitute the known values of v, a, and t into Equation (2): ( ) x = ( +4.50 m/s )( 3.00 s ) − 1 −2.70 m/s2 ( 3.00 s ) = +25.7 m
2
2 Note: Equation (2) can also be obtained by combining Equation (1) with Equation 2.7 x = 1 ( v0 + v ) t , or, with more effort, by combining Equation (1) with Equation 2.9
2 ( v2 = v02 + 2ax ) . 40. REASONING AND SOLUTION As the plane decelerates through the intersection, it
covers a total distance equal to the length of the plane plus the width of the intersection, so
x = 59.7 m + 25.0 m = 84.7 m
The speed of the plane as it enters the intersection can be found from Equation 2.9. Solving
Equation 2.9 for v0 gives
v0 = v 2 − 2ax = (45.0 m) 2 − 2(−5.70 m/s 2 )(84.7 m) = 54.7 m/s 66 KINEMATICS IN ONE DIMENSION The time required to traverse the intersection can then be found from Equation 2.4. Solving
Equation 2.4 for t gives
v − v0 45.0 m/s − 54.7 m/s
= 1.7 s
a
−5.70 m/s 2
______________________________________________________________________________
t= 41. SSM REASONING = As the train passes through the crossing, its motion is described by 1
Equations 2.4 (v = v0 + at) and 2.7 x = (v + v0 )t , which can be rearranged to give
2 v − v0 = at...
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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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