The deceleration of a car traveling on a stra | vedclass

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(A) Given deceleration $\omega = -\frac{dv}{dt} = a \sqrt{v}$.$1$. To find time $t$:$\frac{dv}{dt} = -a \sqrt{v} \implies \int_{v_0}^{0} v^{-1/2} dv =

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$\frac{2 v_0^{3/2}}{3 a}, \frac{2 \sqrt{v_0}}{a}$

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(A) Given deceleration $\omega = -\frac{dv}{dt} = a \sqrt{v}$.$1$. To find time $t$:$\frac{dv}{dt} = -a \sqrt{v} \implies \int_{v_0}^{0} v^{-1/2} dv = \int_{0}^{t} -a dt$$[2 \sqrt{v}]_{v_0}^{0} = -at \implies -2 \sqrt{v_0} = -at \implies t = \frac{2 \sqrt{v_0}}{a}$.$2$. To find distance $s$:Using $\omega = v \frac{dv}{ds} = a \sqrt{v} \implies v \frac{dv}{ds} = -a \sqrt{v} \implies \sqrt{v} dv = -a ds$.Integrating both sides: $\int_{v_0}^{0} v^{1/2} dv = \int_{0}^{s} -a ds$$[\frac{2}{3} v^{3/2}]_{v_0}^{0} = -as \implies -\frac{2}{3} v_0^{3/2} = -as \implies s = \frac{2 v_0^{3/2}}{3 a}$.

The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $\omega = a \sqrt{v}$, where $a$ is a constant. If the initial velocity of the car is $v_0$, the distance the car will travel and the time it takes before it stops are:

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