A car travels with a uniform velocity of $25$ m/s for $5$ s. The brakes are then applied and the car is uniformly retarded and comes to rest in further $10$ s.
Find:
a) the retardation and
b) the distance travelled by the car after applying the brakes.

Graphical representation of the problem:
Initially the car moves with uniform velocity of $25$ m/s for $5$ sec, then brakes are applied and the car comes to rest after $10$ sec i.e. at t $=15$ sec.

Finding retardation:
As car comes to rest from speed of $25\text{m}/\text{s}$ in $10\text{sec}$.
We know that
$\text{Acceleration}\left(a\right)=\frac{\text{Change in velocity}}{\text{time}}\Rightarrow a=\frac{0-25}{10}=-2.5\text{m}/{\text{s}}^2$

So, retardation of the car is $2.5\text{m}/{\text{s}}^2$.

Covered distance after applying the brakes:
We have the equation of motion,
$v^2=u^2+2aS\Rightarrow S=\frac{v^2-u^2}{2a}a=-2.5m{s}^{-2}\text{(Calculated in part a)}\Rightarrow S=\frac{0^2-(25{)}^2}{2\times -2.5}m\Rightarrow S=\frac{625}{5}m=125\text{m}$

Hence, the distance travelled after applying the break is $125\text{m}$.