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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line3 Marks
Question
The acceleration experienced by a boat, after its engine is cut off, given by, $\frac{\text{dv}}{\text{dt}}=-\text{kv}^3,$ where k is a constant. If $v_0$ is the magnitude of velocity at cut off $(t = 0)$, find the magnitude of the velocity at a time t after the cut off.
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Answer

$\frac{\text{dv}}{\text{dt}}=-\text{kv}^3$ Integrating both sides, we get $\int\frac{\text{dv}}{\text{v}}=-\text{k}\int\text{dt}$ $-\frac{1}{2\text{v}^2}=-\text{kt}+\text{c}$
At $t = 0, v = v_0 \therefore \text{c}=-\frac{1}{2\text{v}^2_0}$
$\therefore -\frac{1}{2\text{v}^2}=-\text{kt}-\frac{1}{2\text{v}^2_0}$
$2\text{v}^2=\frac{2\text{v}^2_0}{(2\text{v}^2_0\text{kt}+1)}\text{s}$
$\text{v}=\sqrt{\frac{\text{v}^2_0}{(2\text{v}^2_0\text{kt}+1)}}$

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