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2.Motion in Straight Line
hard
The deceleration experienced by a moving motor boat, after its engine is cut-off is given by $\frac{{dv}}{{dt}} = - k{v^3}$ , where $k$ is constant. If $v_0$ is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time $t$ after the cut-off is
A$\frac {v_0}{2}$
B$v_0$
C${v_0}{e^{ - kt}}$
D$\frac{{{v_0}}}{{\sqrt {\left( {2v_0^2kt + 1} \right)} }}$
Solution
Here, $\frac{d v}{d t}=-k v^{3}$
or $\quad \frac{d v}{v^{3}}=-k d t$ or $\int_{v_{0}}^{v} \frac{d v}{v^{3}}=\int_{0}^{t}-k d t$
$\left[-\frac{1}{2 v^{2}}\right]_{v_{0}}^{v}=-k t$
$\text { or } \quad \frac{1}{2 v_{0}^{2}}-\frac{1}{2 v^{2}}=-k t$
$\therefore \quad \frac{1}{2 v^{2}}=\frac{1+2 v_{0}^{2} k t}{2 v_{0}^{2}}$ or $v^{2}=\frac{v_{0}^{2}}{1+2 v_{0}^{2} k t}$
$\therefore \quad v=\frac{v_{0}}{\sqrt{\left(2 v_{0}^{2} k t+1\right)}}$
or $\quad \frac{d v}{v^{3}}=-k d t$ or $\int_{v_{0}}^{v} \frac{d v}{v^{3}}=\int_{0}^{t}-k d t$
$\left[-\frac{1}{2 v^{2}}\right]_{v_{0}}^{v}=-k t$
$\text { or } \quad \frac{1}{2 v_{0}^{2}}-\frac{1}{2 v^{2}}=-k t$
$\therefore \quad \frac{1}{2 v^{2}}=\frac{1+2 v_{0}^{2} k t}{2 v_{0}^{2}}$ or $v^{2}=\frac{v_{0}^{2}}{1+2 v_{0}^{2} k t}$
$\therefore \quad v=\frac{v_{0}}{\sqrt{\left(2 v_{0}^{2} k t+1\right)}}$
Standard 11
Physics
