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Answer 1

Answer:

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

Answer 2

Answer:

it is -5 m/s..........

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