Mathematics

A disc, sliding on an inclined plane , is found to have its position (measured from the top of the plane) at any instant given by $$x = 3{t^2} + 1$$ where $$x$$ is in meter and $$t$$ in second. Its average velocity in the time interval between $$2s\,to\,2$$is


ANSWER

$$12.3\,m{s^{ - 1}}$$


SOLUTION
$$\begin{array}{l}\\x = 3{t^2} + 1\\v = \frac{{dx}}{{dt}} = \frac{{d\left( {3{t^2} + 1} \right)}}{{dt}}\\average\,velocity = \frac{{\int\limits_2^{2 - 1} {vdt} }}{{\int\limits_2^{2 - 1} {dt} }}\\{V_{avg}} = \frac{{\int\limits_2^{2 - 1} {6tdt} }}{{\int\limits_2^{2 - 1} {dt} }}\\ = 3\frac{{\left( {{{2.1}^2} - {2^2}} \right)}}{{0.1}}\\ = 12.3m{s^{ - 1}}\end{array}$$
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Single Correct Medium Published on 17th 09, 2020
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