posted by Abbie on .
An automobile moving at a constant velocity of 15m/s passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at a constant rate of 2m/s^2 in the same direction. How soon does the second autombile overtake the first?
Let t =0 be the time after the first can passes by.
Solve for t when
15 t = (1/2)*(2.0)*(t-2)^2
The term on the right is the distance covered by the accelerating car that starts 2 seconds late.
15t = (t-2)^2 = t^2 -4t + 4
t^2 -19t +4 = 0
t = [19 +sqrt(345)]/2 = 18.8 seconds
They both will be 282 m from the gas station at that time.