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

We know that the initial relative speed is $v_i=80-25=55\frac{Km}{h}=15.28\frac{m}{s}$ Now, we can find the final relative speed as follows $v_f^2=v_i^2-2ax$ We plug in the known values to obtain: $v_f^2=(15.28)^2+(2)(-2.1)(50)=23.41$ $v_f=4.8\frac{m}{s}$