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.

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Solution

The correct option is D

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At the instant when car B decides to overtake car A, the velocities of cars are:
$v_{A} = 36 \times \frac{5}{18} = 10 m / s, v_{B} = 54 \times \frac{5}{18} = 15 m / s$

And $v_{C} = - 54 \times \frac{5}{18} = - 15 m / s$

Velocity of car B relative to A, $_{C A} =_{C} -_{A} = 15 - 10 = - 25 m / s$

Time the car C requires to just cross A= $\frac{1000}{v_{C A}} = \frac{1000}{25} = 40 s$
In order to avoid accident, car B must overtake A in this time.

So, $1000 = v_{B A} t + \frac{1}{2} a_{B A} t^{2}, 1000 = 5 \times 40 + \frac{1}{2} a_{B A} \times 40^{2}, a_{B A} = 1 m / s^{2}$

Thus the minimum accerleration that car B require to avoid an accident is 1 $m / s^{2}$

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