Question

velocity of a particle is ✓180-4x where x is position then what is acceleration of particle??​

Answer 1

Correct Question

Velocity of a particle is given by the following relation,find the acceleration of the particle :

$\tt \: v(x) = \sqrt{180 - 4x} \: \: {ms}^{ - 1}$

Solution

The acceleration of the particle is 0.5 m/s²

Given

Velocity of the particle is given as :

$\tt \: v(x) = \sqrt{180 - 4x} \: \: {ms}^{ - 1}$

We Know that,

$\tt \: a = \dfrac{dv}{dt} - - - - - - - (1)$

Also $\tt \: t = \dfrac{x}{v}$

Putting t in equation (1),

$\implies \: \tt \: a = \dfrac{dv}{d( \frac{x}{v} )} \\ \\ \implies \: \boxed{ \boxed{ \tt \: a = v \: \dfrac{dv}{dx} }}$

Now,

$\sf \: a \: = v \times \dfrac{dv}{dx} \\ \\ \longrightarrow \: \sf \: \: a = \sqrt{180 - 4x} \times \dfrac{d( \sqrt{180 - 4x} )}{dx} \\ \\ \longrightarrow \: \sf \: a = \sqrt{180 - 4x} \times \dfrac{d(180 - 4x) {}^{ \frac{1}{2} } }{dx} \\ \\ \longrightarrow \: \sf \: a = \cancel{\sqrt{180 - 4x}} \times \dfrac{1}{2 \cancel{( \sqrt{180 - 4x})}} \times \dfrac{d(180 - 4x)}{dx} \\ \\ \huge{ \longrightarrow \: \boxed{ \boxed{\sf \: a = - 2 {ms}^{ - 2} }} } \:$

Answer 2

Given,

Velocity of a particle is related to its position as,

$\blue{ \boxed{ \pink{v = \sqrt{180 - 4x} }}}$

It can also be written as,

$180 - {v}^{2} = 4x$

Acceleration is the rate of change of velocity with respect to time.

$\boxed{ \boxed{a = \frac{dv}{dt}}}$

So, Acceleration of the particle is,

$\displaystyle \frac{dv}{dt} = \frac{d}{dt} ( \sqrt{180 - 4x} ) \\ \\ a = \frac{1}{2 \sqrt{180 - 4x} } \times \frac{d}{dt} (180 - 4x) \\ \\ a = \frac{1}{2 \sqrt{180 - 4x} } \times - 4( \frac{dx}{dt} ) \\ \\ a = \frac{1}{2 \sqrt{180 - 4x} } \times - 4v \\ \\ a = - 2 \times \frac{ \sqrt{180 - 4x} }{ \sqrt{180 - 4x} } \\ \\ a = - 2m {s}^{ - 2}$

Therefore, The acceleration of the particle is - 2m/s².