Question 208: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

We know that $$F 	imes v = $$ Power
According to the question, power is constant.
$$	herefore$$ $$F 	imes v = c\,\,\,\,$$ where $$c=$$ constant
$$	herefore$$ $$m{{dv} \over {dt}} 	imes v = c$$ $$\,\,\,\,\left( \, ight.$$ $$	herefore$$ $$\left. {F = ma = {{mdv} \over {dt}}\,\,} ight)$$
$$	herefore$$ $$m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$$ $$	herefore$$ $${1 \over 2}m{v^2} = ct$$
$$	herefore$$ $$v = \sqrt {{{2c} \over m}}  	imes {t^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
$${{dx} \over {dt}} = \sqrt {{{2c} \over m}}  	imes {t^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}\,\,\,\,$$ where $$v = {{dx} \over {dt}}$$
$$	herefore$$ $$\int\limits_0^x {dx = \sqrt {{{2c} \over m}} }  	imes \int\limits_0^t {{t^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$$
$$x = \sqrt {{{2c} \over m}}  	imes {{2{t^{{aise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{aise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$

Answer

$${t^{3/4}}$$

Answer

$${t^{3/2}}$$

Answer

$${t^{1/4}}$$

Answer

$${t^{1/2}}$$

Work, Energy and Power & Circular Motion

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

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