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2.Motion in Straight Line
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An object, moving with a speed of $6.25\ m/s$, is decelerated at a rate given by:
$\frac{{dv}}{{dt}} = - 2.5\sqrt v $ where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be........$s$
A$8 $
B$1 $
C$2$
D$4$
(AIEEE-2011)
Solution
$\begin{array}{l} \frac{{dv}}{{dt}} = – 2.5\sqrt v \Rightarrow \frac{{dv}}{{\sqrt v }} = – 2.5dt\\ Integrating,\,\int_{6.25}^0 {{v^{{\raise0.5ex\hbox{$\scriptstyle { – 1}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}dv = – 2.5\int_0^t {dt} } \\ \Rightarrow \left[ {\frac{{v{\raise0.5ex\hbox{$\scriptstyle { + 1}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}{{\left( {{\raise0.5ex\hbox{$\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$}}}} = – 2.5t \Rightarrow t = 2\,\sec \end{array}$
Standard 11
Physics
