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A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct^2 (a, b > 0). What will be the nature of motion of the particle when c > 0?(a) Uniform retardation(b) Uniform speed(c) Uniform positive acceleration(d) Uniform velocityThis question was addressed to me in unit test.This intriguing question comes from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

Answer»

Right answer is (c) Uniform POSITIVE acceleration

Explanation: We have, X = a + bt + ct^2……….(1)

Let, v and F be the velocity and acceleration of a PARTICLE at time t seconds.

Then, v = dx/dt = d(a + bt + ct^2)/dt= b + ct……….(2)

And f = dv/dt = d(b + ct)/dt = c……….(3)

CLEARLY, when c > 0,implies f > 0.

Hence in this case the particle moves withan uniform positive acceleration.


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