Unformatted text preview: 30.0 s 2 = 0 . Using the quadratic formula to solve for the time gives t= −8.00 s ± ( 8.00 s ) 2 − 4 ( 4.90) ( −30.0 s 2 )
.
2 ( 4.90) Since the time when the ball reaches the ground must be positive, we use only the positive
solution to find t = 1.79 s .
2.53 During the 0.600 s required for the rig to pass completely onto the bridge, the front
bumper of the tractor moves a distance equal to the length of the rig at constant velocity
of v = 100 k m h . Therefor the length of the rig is ⎡
km ⎛ 0.278 m s ⎞ ⎤
Lrig = v t = ⎢100
⎥ ( 0.600 s ) = 16.7 m .
h ⎜ 1 km h ⎟ ⎥
⎝
⎠⎦
⎢
⎣
While some part of the rig is on the bridge, the front bumper moves a distance
∆x = Lbridge + Lrig = 400 m + 16.7 m . With a constant velocity of v = 100 k m h , the time for
this to occur is t= 2.54 Lbridge + Lrig
v = 400 m + 16.7 m ⎛ 1 km h ⎞
= 15.0 s .
100 k m h ⎜ 0.278 m s ⎟
⎝
⎠ ( ) 1
1
(a) From ∆x = v i t + at 2 , we have 100 m = ( 30.0 m s ) t + −3.50 m s 2 t 2 .
2
2 ( ) This reduces to 3.50 t 2 + ( − 60.0 s ) t + 200 s 2 = 0 , and the quadratic formula gives t= − ( −60.0 s ) ± ( −60.0 s ) 2 − 4 ( 3.50) ( 200 s 2 )
.
2 ( 3.50) The desired time is the smaller solution of t = 4.53 s . The larger solution of
t = 12 .6 s is the time when the boat would pass the buoy moving backwards, assuming it maintained a constant acceleration. 40 CHAPTER 2 (b) The velocity of the boat when it first reaches the buoy is ( ) v f = v i + at = 30.0 m s + − 3.50 m s 2 ( 4.53 s ) = 14.1 m s .
2.55 (a) The acceleration of the bullet is ( 300 m / s ) 2 − ( 400 m / s ) 2
=
=
a=
2 ( ∆x )
2 ( 0.100 m )
v 2 − v i2
f − 3.50 × 105 m s 2 . (b) The time of contact with the board is t= 2.56 v f − vi
a = ( 300 − 400) m s
− 3.50 × 10 m s
5 2 −4
= 2 .86 × 10 s . We assume that the bullet is a cylinder which slows down just as the front end pushes
apart wood fibers.
(a) The acceleration is
a= v 2 − v i2
f
2 ( ∆x ) ( 280 = m s ) − ( 420 m s )
2 2 ( 0.100...
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 Spring '05
 Dr.Ha
 Physics, Acceleration, Velocity

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