A particle is moving in a straight line such that the distance described ‘s’ and the time taken ‘$t$’ are given by $t = as^{2} + bs + c,a 0.$ If $v$ is the velocity of the particle at any time $t$, then acceleration is

Question

A particle is moving in a straight line such that the distance described ‘s’ and the time taken ‘$t$’ are given by $t = as^{2} + bs + c,a > 0.$ If $v$ is the velocity of the particle at any time $t$, then acceleration is

Dr. Abubacker · Accepted Answer

(c)
Given , $t = as^{2} + bs + c$
$\Rightarrow \ \ 1 = 2as\frac{ds}{dt} + b\frac{ds}{dt}$ [differentiating]
$\Rightarrow \ \ 1 = 2asv + bv$ …(i)
$\Rightarrow 0 = 2a\frac{ds}{dt}v + 2as\frac{dv}{dt} + b\frac{dv}{dt}\$ [differentiating]
$\Rightarrow \ \ \ \frac{dv}{dt}(2as + b) = - 2av^{2}$
$\Rightarrow \ \ \ \frac{dv}{dt}\left( \frac{1}{v} \right) = - 2av^{2}$ [from Eq.(i)]
$\Rightarrow \ \ \ \frac{dv}{dt} = - 2av^{3}$

Q. A particle is moving in a straight line such that the distance described ‘s’ and the time taken ‘$t$’ are given by $t = as^{2} + bs + c,a > 0.$ If $v$ is the velocity of the particle at any time $t$, then acceleration is

Explanation

Reviewed by Dr. Abubacker

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