Question

A car of mass 1000kg is moving at a speed of 30m/s. Brakes are applied to bring the car to rest. If the net retarding force is 5000N, the car comes to a stop after travelling ‘d’ m in t sec. then
A. d = 150, t = 5
B. d = 120, t = 8
C. d = 180, t = 6
D. d = 90, t = 6

Answer 1

Hint: In this question, we are given a car which is moving at a speed of 30 m/s. we have to find the distance and time at which the car stops when we apply the retarding force. First we find the acceleration by using the formula $f = ma$. Then we use the first and third equation of motion to find out the distance and time.

Formula used:
The formula of first equation of motion and third equation of motion :-
$v = u + at$
And, ${{v}^{2}}-{{u}^{2}}=2as$
Here, $v$ = Final velocity
$u$ = Initial velocity
$a$ = Acceleration
$t$ = time

Complete step by step solution:
Given a car of mass 1000 kg is moving at a speed of 30 m/s. And the net retarding force is 5000 N. We have to find out the car comes to a stop after travelling d m in t seconds.
We know retarding force $F = ma$
As the force applies is opposite to the motion, then;
$a = \dfrac{-5000}{1000}$= - 5 m/\[{{s}^{2}}\]
when the car stops, ${{v}_{\delta }}=0$

We know according to newton’s first equation of motion,
${{v}_{\delta }}-u=at$
Then $0 – 30 = -5t$
That is t = $\dfrac{-30}{-5}$= 6 sec

Now we know the third equation of motion,
${{v}^{2}}_{\delta }-{{u}^{2}}=2ad$
That is $0-{{30}^{2}}=-2\times 5d$
Then $d=\dfrac{-30\times 30}{-2\times 5}$= 90 m
It means the car comes to a stop after travelling 90 m in 6 seconds.

Thus, option D is correct.

Note: Retarding force is the force that resists relative motion. Sometimes they are also called just retarding forces. Whether the forces actually resist motion depends on who’s looking at a particular situation.