MCQ Solution: Kinematics
A bus moving on a level road with a velocity $$V$$ can be stopped at a distance of $$x$$, by the application of a retarding force $$F$$. The load on the bus is increased by $$25\%$$ by boarding the passengers. Now, if the bus is moving with the same speed and if the same retarding force is applied, the distance travelled by the bus before it stops is:
1) x
2) 1.25x
3) 5x
4) 2.5x
Solution
Let the mass of the system be $$m$$
Initial velocity of bus $$u = V$$
Final velocity of bus $$v = 0$$
Acceleration of the bus $$a =- \dfrac{F}{m}$$
Using $$v^2-u^2 = 2aS$$
$$\therefore$$ $$0-V^2 = 2 \times \bigg( \dfrac{-F}{m} \bigg) \times x$$ $$\implies$$ $$V^2 = \dfrac{2Fx}{m}$$ ...........(1)
Now the mass of the system is increased by $$25$$ %
$$\therefore$$ New mass of the system $$M = 1.25 m$$
Acceleration of bus $$a' = -\dfrac{F}{1.25m}$$
Using $$v^2-u^2 = 2a'x'$$
$$\therefore$$ $$0-V^2 = 2 \times \bigg( \dfrac{-F}{1.25m} \bigg) \times x'$$ $$\implies$$ $$V^2 = \dfrac{2Fx'}{1.25m}$$ .....(2)
From (1) and (2) we get $$ \dfrac{2Fx'}{1.25m}$$ $$ = \dfrac{2Fx}{m}$$
$$\implies $$ $$x' = 1.25 x$$
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