3.8 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?

0 3 Views | Posted 5 months ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    5 months ago

    3.8

    Speed of car A, Va = 36 kmph = 10m/s

    Speed of car B, Vb = 54 kmph = 15 m/s

    Speed of car C, Vc = -54 kmph = -15 m/s

    Relative speed of A, w.r.t. C = Va – (-Vb) = 10 +15 = 25 m/s

    Relative speed of B, w.r.t. A = Vb-Va = 15-10 = 5 m/s

    Distance between AB & AC = 1 km = 1000m

    Time taken by the car C to cover the distance AC, from s = ut

    We get t = 1000/25 = 40s

    To avoid collision with C, the minimum acceleration the vehicle B must have, from the relation, s = ut + (1/2) ft2

    We get 1000 = 5 * 40 + (1/2) f * 402 f = 1 m/s2

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