Velocity of the particle is given as v = [tex]\sqrt{90 - 10x}\[/tex] find its acceleration
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Velocity of the particle is given as v = [tex]\sqrt{90 - 10x}\[/tex] find its acceleration
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Answer :
Velocity of the particle is given by
We have to find acceleration of the particle.
Given : [tex]\tt{v=\sqrt{90-10x}}[/tex]
Squaring both sides, we get
[tex]:\implies\tt\:v^2=90-10x[/tex]
Differentiating both sides wrt t, we get
[tex]:\implies\tt\:2v\:\dfrac{dv}{dt}=\dfrac{90}{dt}-\dfrac{10x}{dt}[/tex]
We know that, differentiation of velocity gives acceleration. Therefore, dv/dt = a
Also differentiation of displacement gives velocity. Therefore, dx/dt = v
[tex]:\implies\tt\:2v\:a=0-10v[/tex]
[tex]:\implies\tt\:2a=-10[/tex]
[tex]:\implies\boxed{\bf{a=-5\:ms^{-2}}}[/tex]
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