JEE MAIN - Physics (2011 - No. 25)
An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by :
$${{dv} \over {dt}} = - 2.5\sqrt v $$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :
$${{dv} \over {dt}} = - 2.5\sqrt v $$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :
2 s
4 s
8 s
1 s
Explanation
Given $${{dv} \over {dt}} = - 2.5\sqrt v $$
$$\Rightarrow {{dv} \over {\sqrt v }} = - 2.5dt$$
On integrating, $$\int_{6.25}^0 {{v^{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \,dv = - 2.5\int_0^t {dt} $$
$$ \Rightarrow \left[ {{{{v^{ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left( {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t$$
$$ \Rightarrow - 2{\left( {6.25} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} = - 2.5t$$
$$ \Rightarrow t = 2\,sec$$
$$\Rightarrow {{dv} \over {\sqrt v }} = - 2.5dt$$
On integrating, $$\int_{6.25}^0 {{v^{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \,dv = - 2.5\int_0^t {dt} $$
$$ \Rightarrow \left[ {{{{v^{ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left( {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t$$
$$ \Rightarrow - 2{\left( {6.25} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} = - 2.5t$$
$$ \Rightarrow t = 2\,sec$$
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