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, v_{A} = 36 km/h = 10 m/s

Velocity of car B, v_{B} = 54 km/h = 15 m/s

Velocity of car C, v_{C} = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,

v_{BA} = v_{B} –(-v_{A} )= 15-(-10) = 25 m/s

Relative velocity of car C with respect to car A,

v_{CA} = v_{C} –v_{A}= 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 2^{nd} equation of motion,

s = ut + 0.5at^{2}

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×40^{2})

∴ a = ms^{-2} = 1 ms^{-2}

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

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