SATHEE: Motion in a Straight Line

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) $3 / 2$

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

(d) (velocity) $2 / 3$

[2010]

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Answer:

Correct Answer: 16. (a)

Solution:

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

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

$∴$ 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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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:

Answer:

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