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]
Show Answer
Answer:
Correct Answer: 16. (a)
Solution:
- (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, $a \propto v^{3 / 2}$
