Posted by john daniel on .
an automobile moving at constant velocity of 15m/sec passes a gasoline station. two seconds later another automobile leaves the gasoline station and accelerates at a constant rate of 2m/ sec squared. how soon will the second automobile overtake the first ?
They will pass when the distances traveled are equal.
15 t = (a/2)(t-2)^2 = (t-2)^2
Solve for t, which is measured from the time the 15 m/s car goes by the station.
t^2 -4t +4 = 15 t
t^2 -19t +4 = 0
t = (1/2)[19 +sqrt345) = 18.79 s
Both cars will have travelled 282 m at that time