An unloaded bus can be stopped by applying brakes on straight road after covering a distance $x$ . Suppose, the passenger add $50 %$ of its weight as the load and the breaking force remains unchanged, how far will the bus go after the application of the brakes? (Velocity of bus in both case is same)

x

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1.5 x

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2 x

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2.5 x

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Solution

The correct option is B $1.5 x$
Same brakes so applied force will be same.
Now as mass becomes (100 +50)%=150% i.e. 1.5 times of initial mass so the retardation will become 1/1.5 times of initial retardation.
If velocity of bus be $u$ then distance travelled before stopping will be given as $S = u^{2} / 2 a$ where $a$ is retardation.
We see that $S \propto 1 / a$ so when retardation a becomes 1/1.5 times then the distance will become inverse of it that it 1.5 times i.e. $1.5 x$

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An unloaded bus can be stopped by applying brakes on straight road after covering a distance x. Suppose, the passenger add 50% of its weight as the load and the breaking force remains unchanged, how far will the bus go after the application of the brakes? (Velocity of bus in both case is same)