Answer
Please see the work below.
Work Step by Step
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
$d=\frac{v_f^2-v_i^2}{2a}$
$d=\frac{(53)^2-(0)^2}{2(7.84)}=180m$
