Motion in a Straight Line - Result Question 16

16. A particle moves a distance $x$ in time t according to equation $x=(t+5)^{-1}$. The acceleration of particle is proportional to:

(a) (velocity) $3 / 2$

(c) $(\text{ distance })^{-2}$

(b) (distance) $)^{2}$

(d) (velocity) $2 / 3$

[2010]

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Answer:

Correct Answer: 16. (a)

Solution:

  1. (a) distance $x=\frac{1}{t+5}$

$\therefore \quad$ velocity $v=\frac{d x}{d t}=\frac{-1}{(t+5)^{2}}$

$\therefore \quad$ acceleration $a=\frac{d^{2} x}{d t^{2}}=\frac{2}{(t+5)^{3}}=2 x^{3}$

Therefore, $v^{3 / 2}=-(t+5)^{-3}$

So, $a \propto v^{3 / 2}$



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