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

We can find the average speed as $v=\frac{140}{3.6}=38.89\frac{m}{s}$ (a) $v=\frac{v_f+ v_i}{2}$ We plug in the known values to obtain: $38.89=\frac{53+v_i}{2}$ $v_i=25\frac{m}{s}$ (b) We know that $a=\frac{\Delta V}{\Delta t}$ We plug in the known values to obtain: $a=\frac{53-25}{3.6}=7.84\frac{m}{s^2}$ Now, we can find the required distance traveled $d=\frac{v_f^2-v_i^2}{2a}$ $d=\frac{(53)^2-(0)^2}{2(7.84)}=180m$