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. $10m/s^2$
• B. $2m/s^2$
• C. $0m/s^2$
• D. $1m/s^2$

Answer 1

The correct option is D

$1m/s^2$

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

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

Velocity of car B relative to A, $\overrightarrow{v_{CA}}=\overrightarrow{v_C}-\overrightarrow{v_A}=15-10=-25m/s$

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

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

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