# A particle moving in a straight line traverses a distance x in time t. If t = x^2/2 + x, then which one is correct?

Category: QuestionsA particle moving in a straight line traverses a distance x in time t. If t = x^2/2 + x, then which one is correct?
Editor">Editor Staff asked 11 months ago

A particle moving in a straight line traverses a distance x in time t. If t = x^2/2 + x, then which one is correct?

(a) The retardation of the particle is the cube of its velocity

(b) The acceleration of the particle is the cube of its velocity

(c) The retardation of the particle is the square of its velocity

(d) The acceleration of the particle is the square of its velocity

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This interesting question is from Calculus Application in chapter Application of Calculus of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

The correct option is (a) The retardation of the particle is the cube of its velocity

The explanation is: We have, t = x^2/2 + x

Therefore, dt/dx = 2x/2 + 1 = x + 1

Thus, if v be the velocity of the particle at time t, then

v = dx/dt = 1/(dt/dx)

= 1/(x + 1) = (x + 1)^-1

Thus dv/dt = d((x + 1)^-1)/dt

= (-1)(x + 1)^-2 d(x + 1)/dt

= -1/(x + 1)^2 * dx/dt

As, 1/(x + 1) = dx/dt,

So, -(dx/dt)^2(dx/dt)

Or dv/dt = -v^2*v   [as, dx/dt = v]

= -v^3

We know, dv/dt = acceleration of a particle.

As, dv/dt is negative, so there is a retardation of the particle.

Thus, the retardation of the particle = -dv/dt = v^3 = cube of the particle.