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
A.
$${t^{\frac{3}{4}}}$$
B.
$${t^{\frac{3}{2}}}$$
C.
$${t^{\frac{1}{4}}}$$
D.
$${t^{\frac{1}{2}}}$$
Answer :
$${t^{\frac{3}{2}}}$$
Solution :
$$\eqalign{
& {\text{We}}\,{\text{knowthat}}\,\,F \times v = {\text{Power}} \cr
& \therefore F \times v = c\,\,{\text{where}}\,c = {\text{constant}} \cr
& \therefore m\frac{{dv}}{{dt}} \times v = c\,\,\left( {\therefore F = ma = \frac{{mdv}}{{dt}}} \right) \cr
& \therefore m\int\limits_0^v v dv = c\int\limits_0^t d t\,\,\therefore \frac{1}{2}m{v^2} = ct \cr
& \therefore v = \sqrt {\frac{{2c}}{m}} \times {t^{\frac{1}{2}}} \cr
& \therefore \frac{{dx}}{{dt}} = \sqrt {\frac{{2c}}{m}} \times {t^{\frac{1}{2}}}\,\,{\text{where}}\,v = \frac{{dx}}{{dt}} \cr
& \int\limits_0^x {dx} = \sqrt {\frac{{2c}}{m}} \times \int\limits_0^t {{t^{\frac{1}{2}}}dt} \cr
& x = \sqrt {\frac{{2c}}{m}} \times \frac{{2{t^{\frac{3}{2}}}}}{3} \Rightarrow x \propto {t^{\frac{3}{2}}} \cr} $$