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According to the equation x=\left(t-5\right)^{-1} , a particle moves a distance x in a time t. The acceleration of a particle is proportional to ?

Option: 1

\left(velocity\right)^\frac{2}{3}


Option: 2

\left(velocity\right)^\frac{3}{2}


Option: 3

\left(distan c e\right)^2


Option: 4

\left(distan c e\right)^{-2}


Answers (1)

best_answer

x=\frac{1}{t+5}  

or  x=(t+5)^{-1}

v=\frac{dx}{dt}=\frac{-1}{\left(t+5\right)^2}

or  v=-(t+5)^{-2}

Now,

Acceleration,

\begin{aligned} & a=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[-(\mathrm{t}+5)^{-2}\right] \\ & \therefore a=\frac{\mathrm{dv}}{\mathrm{dt}}=-(-2)(\mathrm{t}+5)^{-3} =2(\mathrm{t}+5)^{-3}\end{aligned}

\begin{aligned} & \Rightarrow \mathrm{a}=2 \times\left\{-(\mathrm{t}+5)^{-2}\right\}\left\{-(\mathrm{t}+5)^{-2 \times \frac{1}{2}}\right\} \\ & \Rightarrow \mathrm{a}=2 v \times v^{\frac{1}{2}} \\ & \therefore \mathrm{a}=2 v^{\frac{3}{2}} \end{aligned}

Or, a\propto\left(velocity\right)^\frac{3}{2}

Posted by

SANGALDEEP SINGH

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