Question

On a two-lane road, car A is travelling with a speed of 36kmh^-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 I km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Answer 1

ν_A = 36kmh^-1

= 36 × \(\frac{5}{18}\)ms^-1 = 10ms^-1

ν_B = ν_C = 54ms^-1

= 54 × \(\frac{5}{18}\)ms^-1 = 15ms^-1

Relative velocity of B w.r.t. A, ν_BA = 5 ms^-1

Relative velocity of C w.r.t.A, ν_CA= 25ms^-1

Time taken by C to cover distance AC

= \(\frac{1000}{25ms^−1}\) = 40s

Now, for B, 1000 = 5 × 40 + \(\frac{1}{2}\)a × 40 × 40

On simplification, a = 1 ms^-2.