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A car is moving with 10ms^{-1}; there is a bus stop driver needs to apply a break 400 m before that to rest the car at the bus stop; now suppose the driver applied the break at the half distance so the car will cross the bus stop with  \sqrt{x}ms^{-1} velocity; the value of x

Option: 1

50

 

 


Option: 2

200


Option: 3

100

 


Option: 4

75


Answers (1)

best_answer

From the equation of motion

v^{2}=u^{2}+2as

0=u^{2}+2as

S=-\frac{u^{2}}{2a}=-\frac{100}{2a}

400=-\frac{100}{2a}

a=-\frac{1}{8}ms^{-1}

Now break applied at a half distance; S=200 m

v^{2}=(10)^{2}+2*(-\frac{1}{8})*200

 v=\sqrt{50}

Posted by

Sanket Gandhi

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