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Posted by **john daniel** on Saturday, December 1, 2012 at 11:10am.

- physics -
**drwls**, Saturday, December 1, 2012 at 1:11pmThey 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

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