An object, moving with a speed of 6.25 m/s, i | vedclass

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Question

$\begin{array}{l} \frac{{dv}}{{dt}} =  - 2.5\sqrt v  \Rightarrow \frac{{dv}}{{\sqrt v }} =  - 2.5dt\ Integrating,\,\int_{6.25}^0 {{v^{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\begin{array}{l} \frac{{dv}}{{dt}} =  - 2.5\sqrt v  \Rightarrow \frac{{dv}}{{\sqrt v }} =  - 2.5dt\ Integrating,\,\int_{6.25}^0 {{v^{{aise0.5ex\hbox{$\scriptstyle { - 1}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}dv =  - 2.5\int_0^t {dt} } \  \Rightarrow \left[ {\frac{{v{aise0.5ex\hbox{$\scriptstyle { + 1}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}{{\left( {{aise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} ight)}}} ight]_{6.25}^0 =  - 2.5\left[ t ight]_0^t\  \Rightarrow \, - 2{\left( {6.25} ight)^{{aise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} =  - 2.5t \Rightarrow t = 2\,\sec  \end{array}$

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........$s$

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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........$s$