Question

On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h-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

Explanation:

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

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

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

Relative velocity of car B with respect to car A,

vBA = vB – vA

= 15 – 10 = 5 m/s

Relative velocity of car C with respect to car A,

vCA = vC – (– vA)

= 15 + 10 = 25 m/s

At a certain instance, both cars B and C are at the same distance from car A i.e.,

s = 1 km = 1000 m

Time taken (t) by car C to cover 1000 m = 1000 / 25 = 40 s

Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.

From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:

s = ut + (1/2)at2

1000 = 5 × 40 + (1/2) × a × (40)2

a = 1600 / 1600 = 1 ms-2

Answer 2

Answer:

See the attachment

Explanation: