Q. 83.7( 116 Votes )

On a two-lane road, car A is travelling with 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 C is required to avoid an accident?

Answer :

Given,


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) = 25 m/s


Relative velocity of car C with respect to car A,


vCA = vC –vA= 15-10 =5 m/s



At a certain time, both cars B and C are at the same distance,


s = 1 km = 1000m


Time taken (t) by car B to just reach car A is,


t = = 40 s


The minimum acceleration (a) required for car C to just beat car B in reaching car A is given by,


From 2nd equation of motion,


s = ut + 0.5at2


where,


u = Initial velocity = 5 m/s


a = Acceleration/Deceleration


s = Distance covered = 1000 m


t = Time = 40 s


1000 = (5×40) + (0.5×a×402)


a = ms-2 = 1 ms-2


Hence, car C require minimum acceleration, a = 1 ms-2 to beat car B.

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