Problem 10 on Motion in a Straight Line

Description:

Problem

On a two-lane road, car A is travelling with 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?

Solution

Assume Car A to be moving East.

Therefore, Car B moves East approaching car A from behind.

Car C moves West approaching car B from front.

Following sign convention is assumed for this problem −

Therefore,

Velocity of car A = +36 km/h = 10 m/s

Velocity of car B = +54 km/h = 15 m/s

Velocity of car C = −54 km/h = −15 m/s

In the frame of reference of Car A −

Velocity of B with respect to A, v_BA = v_B − v_A = 15 − (+10) = 5 m/s

Velocity of C with respect to A, v_CA = v_C − v_A = −15 − (+10) = −25 m/s

Assume car B moves with an acceleration of a_B, to overtake A.

Therefore, a_BA = a_B - a_A = a_B - 0 = a_B

Visualizing the Problem in the Frame of Car A

To avoid the accident, B has to reach A before C does. We will calculate the bare minimum acceleration of Car B so that it reaches A simultaneously with Car C. Any acceleration higher than that will make Car B reach Car A first and hence, avoid the accident.

Therefore,

Time taken by C to reach A, t = 1000/25 = 40 s (Calculated in the Frame of Car A)

Same time is available for B to overtake A.

Applying, S = ut + 1/2 at², (Equation applied in the Frame of Car A)

1000 = 5 × 40 + 1/2 × a_B × (40)²

a_B = 1 m/s²