Question

A particle moves in a horizontal straight line under retardation kv^3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?

(a) 1/v = 1/u + kx

(b) 1/v = 1/u – 2kx

(c) 1/v = 1/u – kx

(d) 1/v = 1/u + 2kx

I got this question in an internship interview.

I want to ask this question from Calculus Application topic in chapter Application of Calculus of Mathematics – Class 12

Answer 1

Correct answer is (a) 1/v = 1/u + kx

The best explanation: Since the particle is moving in a straight line under a retardation kv^3, hence, we have,

dv/dt = -kv^3   ……….(1)

Or dv/dx*dx/dt = -kv^3

Or v(dv/dx) = -kv^3  [as, dx/dt = v]

Or ∫v^-2 dv = -k∫dx

Or v^-2+1/(-2 + 1) = -kx – B, where B is a integration constant

Or 1/v = kx + B    ……….(3)

Given, v = u, when x = 0; hence, from (3) we get, B = 1/u

Thus, putting B = 1/u in (3) we get,

1/v = 1/u + kx.