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Question

On a two-lane road, car A is travelling with a speed of 36 km/hr. Two cars, B and C approach car A in opposite directions with a speed of 54 km/h. At a certain instant, when the distance AB is equal to AC, both 1 km, B decided to overtake A before C does. What minimum acceleration of car B is required to avoid an accident.


A

10m/s2

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B

2m/s2

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C

0m/s2

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D

1m/s2

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Solution

The correct option is D

1m/s2


\

At the instant when car B decides to overtake car A, the velocities of cars are:
vA=36×518=10 m/s, vB=54×518=15 m/s

And vC=54×518=15 m/s

Velocity of car B relative to A, vCA=vCvA=1510=25m/s

Time the car C requires to just cross A=1000vCA=100025=40s
In order to avoid accident, car B must overtake A in this time.

So, 1000=vBAt+12aBAt2, 1000=5×40+12 aBA×402, aBA=1m/s2

Thus the minimum accerleration that car B require to avoid an accident is 1 m/s2


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