Question

A car is travelling with a speed of $36km/h$. The driver applies the brakes and retards the car uniformly. The car stopped at 5 s. Find
(i) The retardation of the car.
(ii) Distance travelled before it is stopped after applying the brake.

Answer 1

Hint: The initial velocity of the car is given. After applying the brake, the car comes to rest within 5 seconds with uniform retardation. Retardation is negative acceleration. By definition of acceleration, we know v = u + at. Therefore we can find retardation. Also, we can find the distance covered by the car from the formula $s = ut + \dfrac{1}{2}a{t^2}$.

Formula used:
If final velocity is v, initial velocity is u, t is time and acceleration is given by a, then
by definition of acceleration, v = u + at
Distance $s = ut + \dfrac{1}{2}a{t^2}$

Complete step by step solution:
Given that, the car is travelling with a speed of 36km/h.
First, convert all the units to SI.
Initial velocity (u) =$36{{ km/h }} = {{ }}\dfrac{{36 \times 1000}}{{3600}}{{ m/s }} = {{ 10 m/s}}$
When the driver applies the brakes the car stops in 5 s with uniform retardation.
Final velocity (v) = 0
Now, we know by definition of acceleration, v = u + at
$\Rightarrow 0 = 10 + a.5$
$\Rightarrow a = \dfrac{{ - 10}}{5} = - 2{{ m/}}{{{s}}^2}$ (the negative sign implies retardation)
Therefore, retardation is $2{{ m/}}{{{s}}^2}$.
Again, we know
$s = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow s = \left( {10 \times 5} \right) + \dfrac{1}{2}\left( { - 2} \right){\left( 5 \right)^2}$
$ \Rightarrow s = 50 - 25$
$ \Rightarrow s = 25{{ m}}$

Hence, the distance covered by the car before stopping is 25 m.

Note: Note that, negative acceleration is called retardation. Acceleration or retardation is defined by the change in velocity per unit time.
Therefore $a = \dfrac{{v - u}}{t}$
$ \Rightarrow v - u = at$
$ \Rightarrow v = u + at$