## Fundamentals of Physics Extended (10th Edition)

$2.5m/s$
The distance traveled by the fast car is equal to the distance of separation plus the distance traveled by the slow car: $vt=2d+v_{s}t$. Solving this equation for $t$, we get $t=\frac{2d}{v-v_{s}}=\frac{2(48m/s)}{(25m/s)-(5m/s)}=4.8s$. The distance traveled by the slow car during that time can be calculated with the equation $x=v_{s}t=(5m/s)(4.8s)=24m$. However, we need to account for the length of the new car and the buffer zone so you need to subtract $12m$ from $24m$: $(24m)-(12m)=12m$. This is the distance traveled by the shock wave. Since it took the shockwave $4.8$ seconds to travel that distance the velocity can be calculated by dividing the distance by the time: $v=\frac{12m}{4.8s}=2.5m/s$.