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Question -

On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars Band
C approach car A in opposite directionswith a speed of 54 km/h each. At a
certain instant, when the distance ABis equal to AC, both being 1 km, B decides
to overtake A before C does. Whatminimum acceleration of car B is required to
avoid an accident?



Answer -

The speed of car A =36 km/h = 36 x (5/8) = 10 m/s

Speed of car B = 54km/h = 54 x (5/18) = 15 m/s

Speed of car C =– 54 km/h = -54 x (5/18) = -15 m/s (negative sign shows B and C are inopposite direction)

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

Relative speed of Bw.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 carC to cover the distance AC, t = 1000/VAC = 1000/ 25 = 40 s

If a is the acceleration,then

s = ut + (1/2) at2

1000 = (5 x 40) +(1/2) a (40) 2

a = (1000 – 200)/ 800= 1 m/s2

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

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