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October 20, 2014

October 20, 2014

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

- Physics -
**drwls**, Tuesday, August 9, 2011 at 6:28amLet 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.

- Physics -
**Anonymous**, Thursday, January 31, 2013 at 10:24amasjklaj;a

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