Unformatted text preview: 0 = v0A + 1 aA t
2 0 = v0B + 1 aBt
2 t= −2v0A t= aA −2v0B
aB The time for each rocket is the same, so that we can equate the two expressions for t, with
the result that
−2v0A −2v0B
v0A v0B
=
or
=
aA
aB
aA
aB
Solving for aB gives aB = aA
v0A v0B = −15 m/s2
( 8600 m/s ) = −22 m/s2
5800 m/s As expected, the magnitude of the acceleration for rocket B is greater than that for rocket A. 60 KINEMATICS IN ONE DIMENSION 33. REASONING The definition of average velocity is given by Equation 2.2 as the
displacement divided by the elapsed time. When the velocity is constant, as it is for car A,
the average velocity is the same as the constant velocity. We note that, since both
displacement and time are the same for each car, this equation gives the same value for car
B’s average velocity and car A’s constant velocity.
Since the acceleration of car B is constant, we know that its average velocity is given by
Equation 2.6 as vB = 1 ( vB + vB0 ) , where vB is the final velocity and vB0 = 0 m/s is the initial
2
velocity (car B starts from rest). Thus, we can use Equation 2.6 to find the final velocity.
Car B’s constant acceleration can be calculated from Equation 2.4 (vB = vB0 + aBt), which is
one of the equations of kinematics and gives the acceleration as [aB = (vB − vB0)/t]. Since
car B starts from rest, we know that vB0 = 0 m/s. Furthermore, t is given. Therefore,
calculation of the acceleration aB requires that we use the value calculated for the final
velocity vB.
SOLUTION
a. According to Equation 2.2, the velocity of car A is the displacement L divided by the
time t. Thus, we obtain
L 460 m
= 2.2 m/s
vA = =
t
210 s
b. The average velocity of car B is given by Equation 2.6 as vB = 1
2 ( vB + vB0 ) , where vB is the final velocity and vB0 is the initial velocity. Solving for the final velocity and using the
fact that car B starts from rest (vB0 = 0 m/s) gives
vB = 2vB − vB0 = 2vB (1) As discussed in the REASONING, the average velocity of car B is equal to the constant
velocity of car A. Substituting this result into Equation (1), we find that
vB = 2vB = 2vA = 2 ( 2.2 m/s ) = 4.4 m/s
c. Solving Equation 2.4 (vB = vB0 + aBt) for the acceleration shows that aB = vB − vB0
t = 4.4 m/s − 0 m/s
= 0.021 m/s 2
210 s Chapter 2 Problems 61 34. REASONING The entering car maintains a constant acceleration of a1 = 6.0 m/s2 from the
time it starts from rest in the pit area until it catches the other car, but it is convenient to
separate its motion into two intervals. During the first interval, lasting t1 = 4.0 s, it
accelerates from rest to the velocity v01 with which it enters the main speedway. This
velocity is found from Equation 2.4 ( v = v0 + at ) , with v0 = 0 m/s, a = a1, t = t1, and v = v01:
v10 = a1t1 (1) The second interval begins when the entering car enters the main speedway with velocity
v01, and ends when it catches up with the other car, which travels with a constant velocity
v02 = 70.0 m/s. Since both cars begin and end the interval sidebyside, they both undergo
the same displacement x during this interval. The displacement of each car is given by ( ) Equation 2.8 x = v0t + 1 at 2 . For the accelerating car, v0 = v10, and a = a1, so
2
x = v10t + 1 a1t 2
2 (2) For the other car, v0 = v02 and a = 0 m/s2, and so Equation 2.8 yields
x = v2 0 t (3) SOLUTION The displacement during the second interval is not required, so equating the
right hand sides of Equations (2) and (3) eliminates x, leaving an equation that may be
solved for the elapsed time t, which is now the only unknown quantity:
v10t + 1 a1t 2 = v20t
2
v10 t + 1 a1t 2 = v20 t
2
1at
21 t= (4) = v20 − v10 2 ( v20 − v10 )
a1 Substituting Equation (1) for v10 into Equation (4), we find that t= 2 ( v20 − a1t1 )
a1 ( 2 70.0 m/s − 6.0 m/s 2
=
6.0 m/s 2 ( ) ) ( 4.0 s ) = 15 s 62 KINEMATICS IN ONE DIMENSION 35. REASONING The drawing shows the two knights, initially separated by the displacement
d, traveling toward each other. At any moment, Sir George’s displacement is xG and that of
Sir Alfred is xA. When they meet, their displacements are the same, so xG = xA. − Starting point
for Sir George + Starting point
for Sir Alfred xG
xA
d According to Equation 2.8, Sir George's displacement as a function of time is
xG = v0,G t + 1 aG t 2 = ( 0 m/s ) t + 1 aG t 2 = 1 aG t 2
2
2
2 (1) where we have used the fact that Sir George starts from rest ( v0,G = 0 m/s ) .
Since Sir Alfred starts from rest at x = d at t = 0 s, we can write his displacement as (again,
employing Equation 2.8)
xA = d + v0,A t + 1 aA t 2 = d + ( 0 m/s ) t + 1 aA t 2 = d + 1 aA t 2
2
2
2
2 Solving Equation 1 for t ( t 2 = 2 xG / aG ) (2) and substituting this expression into Equation 2 yields 2x
xA = d + 1 aA G
2 aG xG = d + aA aG Noting that xA = xG when the two riders collide, we see that Equation 3 becomes
x xG = d + aA G aG Solving this equation for xG gives xG = d
.
aA
1−
aG (3) Chapter 2 Problems ( 63 ) SOLUTION Sir George’s acceleration is positive aG = +0.300 m/s2 since he starts from
rest and moves to the right (the...
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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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