Question

A car running at $30km/hr$ stops after travelling $8m$ distance on applying brakes. If the same car is running at $60km/hr$it stops after travelling how much distance on applying brakes?

Answer 1

Hint: Use an equation of motion to solve the question . The acceleration of the car will be the same in both cases. The acceleration is defined as the rate of change of velocity . Thus the acceleration of the car will be the same when the speed of the car is doubled.

Complete step by step answer
Initially the car is travelling at a speed of $30km/hr$ ,
So , $u = 30km/hr$
$u = 30 \times \dfrac{5}{{18}}m/s$
$u = \dfrac{{25}}{3}m/s$
$u = $Initial Velocity
Initial Velocity=It is velocity of an object before the acceleration causes a change in the velocity.
Since the car is stopping at the end, the final velocity will be $0$.
So,$v = 0km/hr$
$v = 0m/s$
Final Velocity= It is velocity of an object after the acceleration causes a change in the velocity
Now , Since the car is covering a distance of $8m$during its entire journey,
So, $s = 8m$
So, Using the second equation of motion,
${v^2} = {u^2} + 2as$
We can find the value of acceleration .
Acceleration= Acceleration of a body is defined as the rate of change of velocity.
Since the car is decreasing , retardation happens,
Thus the value of acceleration is negative.
So, Now the formula becomes
$ \Rightarrow {v^2} = {u^2} - 2as$
So , putting the value of $u,v$ and $s$ in the above equation we get the value of $a$as
$ \Rightarrow {0^2} = {\left( {\dfrac{{25}}{3}} \right)^2} - 2 \times a \times 8$
$ \Rightarrow a = 4.34m/{s^2}$
$CASE - 2$
Now we have the initial velocity as $60km/hr$
Let us denote this as $u'$
So , $u' = 60km/hr$
$ \Rightarrow u' = 60 \times \dfrac{5}{{18}}m/s$
$ \Rightarrow u' = \dfrac{{50}}{3}m/s$
Here also, the car stops at the end, so the final velocity(denoted as $v'$)
So , $v' = 0km/hr = 0m/s$
Acceleration of the car is same in both the cases,
$ \Rightarrow a = 4.34m/{s^2}$
So , Using the formula
$ \Rightarrow {v^2} = {u^2} + 2as$
We can find the distance covered.
Here also retardation is happening,
$ \Rightarrow a = - 4.34m/{s^2}$
So , putting the value of $u',v'$ and $a$ in the above equation we get the value of $s$as
$ \Rightarrow {0^2} = {\left( {\dfrac{{50}}{3}} \right)^2} - 2 \times 4.34 \times s$
$ \Rightarrow s = 32m$

Thus the car will stop after covering a distance of $32m$

Note
Since the car stops at the end , retardation is happening and thus the value of acceleration is taken as negative.
Retardation is the negative of acceleration , since the rate of change of velocity is negative in this case.