A point particle moves along a straight line such that x - t where t is time- Then ratio of acceleration to cube of the velocity isA- -3B- -2C- -1D- -0-5

The correct option is B 2Velocity v=dxdt=d (√(t))dt=12√(t) ⇒ v^3=18(√(t))^3 Acceleration a=dvdt=d ( 12×1√(t))dt ⇒ a=12( 12 t^ 3/2) ⇒ a=( 141(√(t))^3)= 2v^3 ⇒ ...

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Standard XII

Mathematics

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A point parti...

Question

A point particle moves along a straight line such that $x = \sqrt{t}$ where $t$ is time. Then ratio of acceleration to cube of the velocity is

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Solution

The correct option is B $- 2$
Velocity
$v = \frac{d x}{d t} = \frac{d (\sqrt{t})}{d t} = \frac{1}{2 \sqrt{t}} \Rightarrow v^{3} = \frac{1}{8 (\sqrt{t})^{3}}$

Acceleration
$a = \frac{d v}{d t} = \frac{d (\frac{1}{2} \times \frac{1}{\sqrt{t}})}{d t} \Rightarrow a = \frac{1}{2} (- \frac{1}{2} t^{- \frac{3}{2}}) \Rightarrow a = (- \frac{1}{4} \frac{1}{(\sqrt{t})^{3}}) = - 2 v^{3} \Rightarrow \frac{a}{v^{3}} = - 2$

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A point particle moves along a straight line such that x=√t where t is time. Then ratio of acceleration to cube of the velocity is

The correct option is B −2Velocity v=dxdt=d (√t)dt=12√t⇒v3=18(√t)3 Acceleration a=dvdt=d (12×1√t)dt⇒a=12(−12t−32)⇒a=(−141(√t)3)=−2v3⇒av3=−2

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