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Find the acceleration and velocity using differentiation if i)x=4/t^3 and ii)x=5t^2+under root tLet me answer plz..with explanation ​

Answer»

\mathbb{\underline {ANSWER}}

(i) X = \frac{4}{t^3}

Acceleration = \frac{48}{t^{5} }

Velocity = \frac{-12}{t^{4}}

(ii) x = 5t² + √t

Acceleration = 10+\frac{1}{4\sqrt{t^3} }

Velocity = 10t+\frac{1}{<klux>2</klux>\sqrt{t} }

\mathbb{\underline {EXPLANATION}}

We know that,

Velocity = Rate of change of displacement

V = dx/dt

Acceleration = Rate of change of velocity

a = dv/dt

1)For the first CASE,

x = \frac{4}{t^3}

Hence,

v = dx/dt

v = \frac{d}{dt} (\frac{4}{t^3} )= \frac{d}{dt} (4t^{-3})= -3 * 4t^{-3-1}=-12t^{-4}=\frac{-12}{t^4}

a = dv/dt

a = \frac{d}{dt} (\frac{-12}{t^4} )=\frac{d}{dt} (-12t^{-4})=-4 * -12t^{-4-1}=48t^{-5}=\frac{48}{t^5}

2)For the SECOND case,

x = 5t² + √t

Hence,

v = dx/dt

v = \frac{d}{dt} (5t^{2} +\sqrt{t} )= \frac{d}{dt} (5t^2)+\frac{d}{dt} (\sqrt{t})=2*5t + \frac{1}{2}t^{\frac{1}{2}-1 } =10t+\frac{1}{2\sqrt{t} }

a = dv/dt

a = \frac{d}{dt} (10t + \frac{1}{2\sqrt{t}} )= \frac{d}{dt} (10t)+\frac{d}{dt} (\frac{1}{2\sqrt{t} } )=10 +\frac{1}{2}* \frac{1}{2} *t^{\frac{-1}{2} -1}=10+\frac{1}{4\sqrt{t^3} }


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