Question

Velocity of the particle is given as v = $\sqrt{90 - 10x}$ find its acceleration

Answer 1

Answer :

Velocity of the particle is given by

• $\bf{v=\sqrt{90-10x}}$

We have to find acceleration of the particle.

Given : $\tt{v=\sqrt{90-10x}}$

Squaring both sides, we get

$:\implies\tt\:v^2=90-10x$

Differentiating both sides wrt t, we get

$:\implies\tt\:2v\:\dfrac{dv}{dt}=\dfrac{90}{dt}-\dfrac{10x}{dt}$

We know that, differentiation of velocity gives acceleration. Therefore, dv/dt = a

Also differentiation of displacement gives velocity. Therefore, dx/dt = v

$:\implies\tt\:2v\:a=0-10v$

$:\implies\tt\:2a=-10$

$:\implies\boxed{\bf{a=-5\:ms^{-2}}}$

Answer 2

Velocity of the particle is given by

\bf{v=\sqrt{90-10x}}v=

90−10x

We have to find acceleration of the particle.

Given : \tt{v=\sqrt{90-10x}}v=

90−10x

Squaring both sides, we get

:\implies\tt\:v^2=90-10x:⟹v

2

=90−10x

Differentiating both sides wrt t, we get

:\implies\tt\:2v\:\dfrac{dv}{dt}=\dfrac{90}{dt}-\dfrac{10x}{dt}:⟹2v

dt

dv

=

dt

90

−

dt

10x

We know that, differentiation of velocity gives acceleration. Therefore, dv/dt = a

Also differentiation of displacement gives velocity. Therefore, dx/dt = v

:\implies\tt\:2v\:a=0-10v:⟹2va=0−10v

:\implies\tt\:2a=-10:⟹2a=−10

:\implies\boxed{\bf{a=-5\:ms^{-2}}}:⟹

a=−5ms

−2