Question

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

Answer 1

The velocity of the car just before the brakes are applied is the initial velocity of the retarding motion of the car. Use the kinematic equation to calculate the deceleration of the car. Then use the kinematic equation to express the distance travelled by the car after the brakes are applied.

Formula used:
$v = u + at$
Here, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
$s = ut + \dfrac{1}{2}a{t^2}$,
where, s is the displacement.

Complete step by step answer:
We have given that the car was travelling with uniform velocity 25 m/s and the brakes are then applied. The car is brought to rest after 10s. Just before the brakes are applied, the velocity of the car is 25 m/s and when the car is brought to rest, its velocity becomes zero. Therefore, in this case, the initial velocity of the car in its decelerated motion is $u = 25\,{\text{m/s}}$ and final velocity is $v = 0\,{\text{m/s}}$.

We can use kinematic equation to determine the acceleration of the car after the brakes are applied as follows,
$v = u + at$
Here, a is the acceleration (deceleration) of the car.
Substituting $u = 25\,{\text{m/s}}$, $v = 0\,{\text{m/s}}$ and $t = 10\,{\text{s}}$, we get,
$0 = 25 + a\left( {10} \right)$
$\Rightarrow a = - 2.5\,{\text{m/}}{{\text{s}}^2}$
The negative sign of the acceleration implies that there is retardation.

We can calculate the distance travelled by the car after the brakes are applied using the kinematic equation as follows,
$s = ut + \dfrac{1}{2}a{t^2}$
Substituting $u = 25\,{\text{m/s}}$, $v = 0\,{\text{m/s}}$, $a = - 2.5\,{\text{m/}}{{\text{s}}^2}$ and $t = 10\,{\text{s}}$, we get,
$s = \left( {25} \right)\left( {10} \right) + \dfrac{1}{2}\left( { - 2.5} \right){\left( {10} \right)^2}$
$\Rightarrow s = 250 - 125$
$\therefore s = 125\,{\text{m}}$

Therefore, the distance travelled by the car after the brakes are applied is 125 m.

Note: Students should never ignore the sign of the acceleration of the body. The negative sign denotes that the velocity of the body is decreasing due to retarding force acting on the body. In this case, the retarding force is the frictional force between the tyres and the road.