A cyclist moving with constant velocity of 6.0 m/s forward passes a car that is just starting. If
the car has constant acceleration of 2.0 m/s 2 , where and when will the car overtake the cyclist?
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Answer:
They are both the same distance from the starting point:
d_cyclist=d_car
vt=(at^2)/2
2v=at
The time is
t=2v/a=2 6.0/2.0=6.0 s.
The distance is
d=vt=(6.0)(6.0)=36 m.
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