Motion in a Straight Line - Result Question 10
10. A particle of unit mass undergoes onedimensional motion such that its velocity varies according to $v(x)=b x^{-2 n}$
where $b$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as the function of $x$, is given by:
(a) $-2 n b^{2} x^{-4 n-1}$
(b) $-2 b^{2} x^{-2 n+1}$
(c) $-2 n b^{2} e^{-4 n+1}$
(d) $-2 n b^{2} x^{-2 n-1}$
[2015]
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Answer:
Correct Answer: 10. (a)
Solution:
- (a) Given, $v(x)=b x^{-2 n}$
$a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}$
$=v \cdot \frac{d v}{d x}$
So, $\frac{d v}{d x}=-2 n b x^{-2 n-1}$
Acceleration of the particle as function of $x$,
$ \begin{aligned} a & =v \frac{d v}{d x}=b x^{-2 n}{b(-2 n) x^{-2 n-1}} \\ & =-2 n b^{2} x^{-4 n-1} \end{aligned} $
For one dimensional motin, the angle between velocity and acceleration is either $0^{\circ}$ or $180^{\circ}$ and it does not change with time.
