JEE MAIN - Physics (2003 - No. 39)
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $$'t'$$ is proportional to
$${t^{3/4}}$$
$${t^{3/2}}$$
$${t^{1/4}}$$
$${t^{1/2}}$$
Explanation
We know that $$F \times v = $$ Power
According to the question, power is constant.
$$\therefore$$ $$F \times v = c\,\,\,\,$$ where $$c=$$ constant
$$\therefore$$ $$m{{dv} \over {dt}} \times v = c$$ $$\,\,\,\,\left( \, \right.$$ $$\therefore$$ $$\left. {F = ma = {{mdv} \over {dt}}\,\,} \right)$$
$$\therefore$$ $$m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$$ $$\therefore$$ $${1 \over 2}m{v^2} = ct$$
$$\therefore$$ $$v = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
$${{dx} \over {dt}} = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}\,\,\,\,$$ where $$v = {{dx} \over {dt}}$$
$$\therefore$$ $$\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$$
$$x = \sqrt {{{2c} \over m}} \times {{2{t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
According to the question, power is constant.
$$\therefore$$ $$F \times v = c\,\,\,\,$$ where $$c=$$ constant
$$\therefore$$ $$m{{dv} \over {dt}} \times v = c$$ $$\,\,\,\,\left( \, \right.$$ $$\therefore$$ $$\left. {F = ma = {{mdv} \over {dt}}\,\,} \right)$$
$$\therefore$$ $$m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$$ $$\therefore$$ $${1 \over 2}m{v^2} = ct$$
$$\therefore$$ $$v = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
$${{dx} \over {dt}} = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}\,\,\,\,$$ where $$v = {{dx} \over {dt}}$$
$$\therefore$$ $$\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$$
$$x = \sqrt {{{2c} \over m}} \times {{2{t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
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