The deceleration experienced by a moving moto | vedclass

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(D) Given the deceleration equation: $\frac{dv}{dt} = -kv^3$.Rearranging the terms to integrate: $\frac{dv}{v^3} = -k dt$.Integrating both sides with

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$\frac{v_0}{\sqrt{2v_0^2kt + 1}}$

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(D) Given the deceleration equation: $\frac{dv}{dt} = -kv^3$.Rearranging the terms to integrate: $\frac{dv}{v^3} = -k dt$.Integrating both sides with limits from $v_0$ to $v$ for velocity and $0$ to $t$ for time:$\int_{v_0}^{v} v^{-3} dv = \int_{0}^{t} -k dt$.Evaluating the integrals:$\left[ \frac{v^{-2}}{-2} \right]_{v_0}^{v} = -kt$.$-\frac{1}{2} \left( \frac{1}{v^2} - \frac{1}{v_0^2} \right) = -kt$.$\frac{1}{v^2} - \frac{1}{v_0^2} = 2kt$.$\frac{1}{v^2} = \frac{1}{v_0^2} + 2kt = \frac{1 + 2v_0^2kt}{v_0^2}$.Taking the reciprocal and square root:$v^2 = \frac{v_0^2}{1 + 2v_0^2kt}$.$v = \frac{v_0}{\sqrt{1 + 2v_0^2kt}}$.

The deceleration experienced by a moving motor boat,after its engine is cut-off,is given by $\frac{dv}{dt} = -kv^3$,where $k$ is a 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:

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