## Physics (10th Edition)

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: $24.0t + 3.0t^{2}=70.0t$ $3.0t^{2}-46.0t=0$ $3.0t(t-\frac{46}{3})=0$ $t=0, t=\frac{46}{3}$ $t=\frac{46}{3}\approx 15.3\; s$