$A$ motor car is moving with a velocity of $108 \, km \, h^{-1}$ and it takes $4 \, s$ to stop after the brakes are applied. Find the force exerted by the brakes on the motor car if its mass along with the passengers is $1000 \, kg$.

(N/A) Mass of the car along with passengers,$m = 1000 \, kg$.
Initial velocity,$u = 108 \, km \, h^{-1} = \frac{108 \times 1000}{3600} \, m \, s^{-1} = 30 \, m \, s^{-1}$.
Final velocity,$v = 0 \, m \, s^{-1}$ (since the car stops).
Time taken,$t = 4 \, s$.
Acceleration,$a = \frac{v - u}{t} = \frac{0 - 30}{4} = -7.5 \, m \, s^{-2}$.
Force exerted by the brakes,$F = m \times a = 1000 \, kg \times (-7.5 \, m \, s^{-2}) = -7500 \, N$.
The negative sign indicates that the force is applied in the direction opposite to the motion of the car.
Therefore,the magnitude of the force exerted by the brakes is $7500 \, N$.