Question

A car with an acceleration of 2m/s^-2 starts from point P at 8:00 a.m. ten seconds later a second car with an acceleration of 3m/s^-2 starts from the same point to overtake the first.

Determine the time when the second car overtakes the first.

Answer 1

$s_1=\frac{a_1t^2}{2},$

$s_2=\frac{a_2(t-10)^2}{2},$

$s_1=s_2,$ $\implies$

$t=\frac{10}{1-\sqrt{\frac{a_1}{a_2}}}\approx54.5~s,$ or

8:00:54.5 a.m.