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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 + 2kxI 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»

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.


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