Question

The velocity v of a particle moving along x-axis varies with position x as v = alpha x , where alpha is constant . What is the graph of acceleration with time ? ​

Answer 1

Answer:

See attachment.

Explanation:

According to Question ,

$\sf \longrightarrow v = \alpha \sqrt{x}$

Differenciate both sides wrt x

$\sf \longrightarrow \dfrac{dv}{dx}= \dfrac{d(\alpha \sqrt{x}}{dx} \\\\\\\sf\longrightarrow \dfrac{dv}{dx}= \alpha \dfrac{dx^{\frac{1}{2}}}{dx} \\\\\\\sf\longrightarrow \dfrac{dv}{dx}= \dfrac{\alpha}{2}x^{\frac{-1}{2}}$

Multiplying both sides by v ,

$\sf\longrightarrow \dfrac{vdv}{dx}= \dfrac{v\alpha}{2}x^{\frac{-1}{2}} \\\\\\\sf\longrightarrow a = \alpha \sqrt{x} \dfrac{\alpha}{2}\dfrac{1}{\sqrt{x}} \\\\\\\sf\longrightarrow\boxed{\sf \red{ accl^n =\dfrac{\alpha^2}{2}}}$

Therefore the graph will be a straight line parallel to the time axis .

Answer 2

$given \\ velocity = v \\ it \: varies \: with \: position \: v = a \sqrt{x} \\ the \: acceleration \: of \: the \: particle \: is \: given \: by \\ a = v \\ v = a \sqrt{x} \\ \frac{dv}{dx} = \frac{a}{2 \times \sqrt{x} } \\ a = a \sqrt{x} \times \frac{a}{2 \sqrt{x} } \\ = \frac{ {a}^{2} }{2}$