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

Ans:

According to the question speed of car A is 36 km/h

or, 36 x (5/8) = 10 m/s

And the speed of car B is 54 km/h

Or, 54 x (5/18) = 15 m/s

And the speed of car C is – 54 km/h

Or, -54 x (5/18) = -15 m/s

Here, negative sign implies that B and C are moving in opposite directions

Now, we will determine the relative speed of A with respect to C

V_AC= V_A – V_B = 10 – (-15)

V_AC  = 25 m/s

Similarly, the relative speed of B with respect to A

V_BA = V_B – V_A = 15 – 10

V_BA = 5 m/s

Now we are given that distance AB = distance AC = 1 km

Or, AB = AC = 1000 m

Then, the time taken by car C to move the distance AC is

t = 1000/V_AC = 1000/ 25

t = 40 s

Let a be the acceleration, then using the expression below –

=> s = ut + (1/2) at²

Substituting values in the above expression, we get

1000 = (5 x 40) + (1/2) a (40) 2

Upon re-arranging =>  a = (1000 – 200)/ 800

a = 1 m/s²

Therefore, the minimum acceleration of car B in order to avoid an accident is 1 m/s²