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

Asked by Pragya Singh | 1 year ago |  63

##### Solution :-

The speed of car A = 36 km/h = 36 x ($$\dfrac{5}{8}$$) = 10 m/s

Speed of car B = 54 km/h = 54 x ($$\dfrac{5}{18}$$) = 15 m/s

Speed of car C = – 54 km/h = -54 x ($$\dfrac{5}{18}$$

= -15 m/s (negative sign shows B and C are in opposite direction)

Relative speed of A w.r.t C, VAC= VA – VB = 10 – (-15) = 25 m/s

Relative speed of B w.r.t A, VBA = VB – VA = 15 – 10 = 5 m/s

Distance between AB = Distance between AC = 1 km = 1000 m

Time taken by the car C to cover the distance AC, t = 1000/VAC = $$\dfrac{1000}{25}$$ = 40 s

If a is the acceleration, then

s = ut + ($$\dfrac{1}{2}$$) at2

1000 = (5 x 40) + ($$\dfrac{1}{2}$$) a (40) 2

a = $$\dfrac{ (1000 – 200)}{800}$$ = 1 m/s2

Thus, the minimum acceleration of car B required to avoid an accident is 1 m/s2

Answered by Abhisek | 1 year ago

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