Question

A particle moves in a straight line and its position
x at time t is given by x2 = 2+ t. Its acceleration
is given by?
answer is -¼x-³. explain how.​

Answer 1

The position of the particle at time $\sf{t}$ is given by,

$\longrightarrow\sf{x^2=t+2}$

Differentiating with respect to $\sf{t,}$

$\longrightarrow\sf{\dfrac{d}{dt}\left(x^2\right)=\dfrac{d}{dt}(t+2)}$

$\longrightarrow\sf{\dfrac{d}{dx}\left(x^2\right)\cdot\dfrac{dx}{dt}=1+0}$

$\longrightarrow\sf{2xv=1}$

$\longrightarrow\sf{v=\dfrac{1}{2}\,x^{-1}\quad\quad\dots(1)}$

Again differentiating with respect to $\sf{t,}$

$\longrightarrow\sf{\dfrac{dv}{dt}=\dfrac{d}{dt}\left(\dfrac{1}{2}\,x^{-1}\right)}$

$\longrightarrow\sf{a=\dfrac{1}{2}\cdot\dfrac{d}{dx}\left(x^{-1}\right)\cdot\dfrac{dx}{dt}}$

$\longrightarrow\sf{a=\dfrac{1}{2}\left(-x^{-2}\right)v}$

From (1),

$\longrightarrow\sf{a=-\dfrac{1}{2}\,x^{-2}\cdot\dfrac{1}{2}\,x^{-1}}$

$\longrightarrow\underline{\underline{\sf{a=-\dfrac{1}{4}\,x^{-3}}}}$

This is the acceleration of the particle.