Answer
15.3 s
Work Step by Step
Speed when entering main speedway:
$v=u+at=(0)+(6.0)(4.0)=24.0\; m/s$
Distance traveled by the car:
$x=ut+\frac{1}{2}at^{2}=(24.0)t + \frac{1}{2}(6.0)t^{2}=24.0t + 3.0t^{2}$
Distance traveled by the other car:
$x=ut+\frac{1}{2}at^{2}=(70.0)t+\frac{1}{2}(0)t^{2}=70.0t$
We now use a system of linear equations:
$24.0t + 3.0t^{2}=70.0t$
$3.0t^{2}-46.0t=0$
$3.0t(t-\frac{46}{3})=0$
$t=0, t=\frac{46}{3}$
$t=\frac{46}{3}\approx 15.3\; s$
