A speeding car is traveling at 92.0 km/hr toward a police car at rest, facing the same direction as the speeding car. If the police car begins accelerating when the speeding car is 250.0 m behind the police car, what must the police car's acceleration be in order for the police car to reach the speeding car's velocity at the moment the speeding car catches up? Assume that the speeding car does not slow down.
10 points to the first to answer it correctly! An explanation would be great as well.
Thanks.
10 points to the first to answer it correctly! An explanation would be great as well.
Thanks.
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92 kph = 25.56 m/s (v )
The police car will accelerate for a distance "d"
d = 1/2 .a . t^2
v = a . t
d = 1/2 . v . t
The speeding car will cover a distance (250 + d )
Time to coincidence = 250 + d / v
2d = vt = v . (250 + d ) /v = 250 + d
d = 250 m
a = v^2 / 2 d = 25.56^2 / 500 = 1.31 m/s/s
The police car will accelerate for a distance "d"
d = 1/2 .a . t^2
v = a . t
d = 1/2 . v . t
The speeding car will cover a distance (250 + d )
Time to coincidence = 250 + d / v
2d = vt = v . (250 + d ) /v = 250 + d
d = 250 m
a = v^2 / 2 d = 25.56^2 / 500 = 1.31 m/s/s