## Essential University Physics: Volume 1 (4th Edition)

We know that the distance traveled in the first fifteen minutes can be determined as $d_1=v_1t_1$ We plug in the known values to obtain: $d_1=(20\frac{mi}{h})(\frac{15}{60})=5mi$ The required distance to be covered is 20 mi in 25 min so $v_2=\frac{20}{\frac{25}{60}}$ $v_2=48\frac{mi}{h}$