Posted by Abbie on Tuesday, August 9, 2011 at 12:13am.
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.
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