Pls solve this one of phusics

Dear student,
Given,
initial velocity u =25 m/s
final velocity v=0
time with constant velocity =5 sec
time related to brake application =10 sec
(i) before the brakes are applied, let us assume the the distance travelled by  the car is S;
(before applying the brakes, there is not external force, so acceleration=0)
$$\begin{gathered} dis\tan ce = speed \times time \\ S= 25\times 5 =125 m \\ \left(ii\right) acceleration =\frac{v-u}{t}=\frac{0-25}{10}=-5/2=-2.5 m/s^2 \end{gathered}$$
as the sign of acceleration is negative, so it denotes the car is retarding. And the retardation = 2.5 m/s²

(iii)
after applying brakes, the time taken to come stop 10 sec
Let assume that S^' be the distance travelled after applying the brakes.
Initial velocity =u= 25 m/s
final velocity v= 0 m/s
using the third equation of motion:
$$\begin{gathered} v^2- u^2= 2as \\ we get, \\ {0}^2-{25}^2= 2(-2.5)S\text{'} \\ S\text{'}=625/5 =125 m \end{gathered}$$

regards,