Question

the acceleration experienced by a car after the engine is switched off is given by dv/dt = -kv^3, where k is a constant. if v• is the magnitude of velocity at switch off, find the magnitude of velocity at time t after switch off? ​

Answer 1

Answer:

$\dfrac{dv}{dt} = -kv^3\\\\\\\text{Transposing v to the LHS we get,}\\\\\dfrac{dv}{v^3} = -k.dt\\\\\text{Integrating on both sides we get:}\\\\\int\limits \dfrac{dv}{v^3} = \int\limits -k.dt\\\\\\\dfrac{-1}{2v^2} = -kt + C$

$\dfrac{1}{v^2} = 2kt + C$

At t = 0, we get v = v₀. Hence the value of constant C is:

$\dfrac{1}{v_0^2} = C$

Hence the equation becomes:

$\dfrac{1}{v^2} = 2kt + \dfrac{1}{v_0^2}\\\\\\\implies \dfrac{1}{v^2} = \dfrac{2ktv_0^2 + 1}{v_0^2}\\\\\\\text{Reciprocating on both sides we get:}\\\\\\\implies v^2 = \dfrac{ v_0^2}{2ktv_0^2+1}\\\\\\\boxed{ v = \sqrt{\dfrac{ v_0^2}{2ktv_0^2+1}}}$

This is the required answer.