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A particle of unit mass undergoes one dimensional motion such that its velocity varies according to $ v(x)= \beta {x^{ - 2n}}$, where $\beta$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$, is given by
$-2n$${\beta ^2}{X^{ - 2n - 1}}$
$-2n$${\beta ^2}{X^{ - 4n - 1}}$
$-2n$${\beta ^2}{X^{ - 2n + 1}}$
$-2n$${\beta ^2}{X^{ - 4n + 1}}$
Solution
$\begin{array}{l}
Accordind\,to\,question,\,velocity\,of\,unit\,\\
mass\,{\rm{varies}}\,as\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v\left( x \right) = \beta {x^{ – 2n}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{dv}}{{dx}} = – 2n\beta {x^{ – 2n – 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)\\
Acceleration\,of\,the\,particle\,is\,give\,by\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}} \times \frac{{dx}}{{dt}} = \frac{{dv}}{{dx}} \times v\\
{\rm{Using}}\,equation\,\left( i \right)\,and\,\left( {ii} \right),\,we\,get\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = \left( { – 2n\beta {x^{ – 2n – 1}}} \right) \times \left( {\beta {x^{ – 2n}}} \right) = – 2n{\beta ^2}{x^{ – 4n – 1}}
\end{array}$
Similar Questions
Velocity of a particle is in negative direction with constant acceleration in positive direction. Then, match the following columns.
| Colum $I$ | Colum $II$ |
| $(A)$ Velocity-time graph | $(p)$ Slope $\rightarrow$ negative |
| $(B)$ Acceleration-time graph | $(q)$ Slope $\rightarrow$ positive |
| $(C)$ Displacement-time graph | $(r)$ Slope $\rightarrow$ zero |
| $(s)$ $\mid$ Slope $\mid \rightarrow$ increasing | |
| $(t)$ $\mid$ Slope $\mid$ $\rightarrow$ decreasing | |
| $(u)$ |Slope| $\rightarrow$ constant |
