Question

On a two-lane road, car A Is travelling with a speed of 36 kmh^-1 . Two cars B and C approach car A in opposite directions with a speed of 54 kmh^-1 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?

Answer 1

V_A= 36 kmh^-1 = 10 ms^-1

VB = Vc = 54 kmh^-1 = 15 ms^-1

Velocity Of B with respect to A

= V_BA = V_B – V_A

= 15m - 10 = 5 ms^-1

Velocity of C with respect to A.

V_CA = V_C – V_A

= 15 – (-10) = 25 ms^-1

S = V_CA t

1000 m = 25 × t

⇒ t = 40 s

C will take 40 s to cover AC. In that time, B must cover AB to overtake and avoid collisions.

s = V_BA t + (1/2) at²

1000 = 5 × 40 + (1/2) × a × 40²

⇒ a = 1 ms^-2

Acceleration of B = 1 ms^-2