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

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(D) Given the deceleration equation: $\frac{dv}{dt} = -2.5\sqrt{v}$.Rearranging the terms to integrate: $\frac{dv}{\sqrt{v}} = -2.5\, dt$.Integrating

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(D) Given the deceleration equation: $\frac{dv}{dt} = -2.5\sqrt{v}$.Rearranging the terms to integrate: $\frac{dv}{\sqrt{v}} = -2.5\, dt$.Integrating both sides with limits from initial speed $v = 6.25\, m/s$ to final speed $v = 0\, m/s$ at time $t$:$\int_{6.25}^{0} v^{-1/2} dv = \int_{0}^{t} -2.5\, dt$.Evaluating the integral: $[2\sqrt{v}]_{6.25}^{0} = -2.5[t]_{0}^{t}$.Substituting the limits: $2(\sqrt{0} - \sqrt{6.25}) = -2.5(t - 0)$.$2(0 - 2.5) = -2.5t$.$-5 = -2.5t$.$t = \frac{5}{2.5} = 2\, s$.

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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