Question 85: Motion in 1D

Question

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 :

Accepted Answer

Given $${{dv} \over {dt}} =  - 2.5\sqrt v  $$
$$\Rightarrow {{dv} \over {\sqrt v }} =  - 2.5dt$$
On integrating, $$\int_{6.25}^0 {{v^{ - {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \,dv =  - 2.5\int_0^t {dt} $$
$$ \Rightarrow \left[ {{{{v^{ + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left( {{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \right)}}} \right]_{6.25}^0 =  - 2.5\left[ t \right]_0^t$$
$$ \Rightarrow  - 2{\left( {6.25} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} =  - 2.5t$$
$$ \Rightarrow t = 2\,sec$$

Answer

2 s

Answer

4 s

Answer

8 s

Answer

1 s

Motion in 1D

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 :

Choose the correct answer: