Question

If he has to continue breaking with the same constant retardation, how much longer would it take for him to stop and how much additional distance he would cover?

Answer 1

Answer is A.
Now with known acceleration, we can find the total time for the car to go from velocity, $u_x=25m/s$ to $v_x=0$ Solving for t, we find
$t\frac{v_x-u_x}{a_x}=\frac{0-25}{-1}$ = 25 s.
The total distance covered is
$x=x_1+u_{x}t+\frac{1}{2}{a}_x{t}^2$
$=0+\left(25\right)\left(25\right)+\frac{1}{2}(-1)(25{)}^2$
= 625 - 312.5
= 312.5 m.
Additional distance covered
= 312.5 - 200
= 112.5 m.
Hence, the time taken for him to stop is 25 s and the additional distance would he cover is 112.5 m.