The acceleration f text{ ft/sec}^2 of a parti | vedclass

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(B, C) Given the acceleration $f = \frac{dv}{dt} = 6 - \sqrt{1.2t}$.At $t = 0$, $v = 0$.The velocity is maximum when the acceleration $f = 0$.Setting

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$T = 30 	ext{ sec}$

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(B, C) Given the acceleration $f = \frac{dv}{dt} = 6 - \sqrt{1.2t}$.At $t = 0$, $v = 0$.The velocity is maximum when the acceleration $f = 0$.Setting $f = 0$, we get $6 - \sqrt{1.2t} = 0 \Rightarrow \sqrt{1.2t} = 6 \Rightarrow 1.2t = 36 \Rightarrow t = \frac{36}{1.2} = 30 	ext{ sec}$.Thus, the time $T = 30 	ext{ sec}$.To find the maximum velocity $v$, we integrate the acceleration with respect to time from $t = 0$ to $t = 30$:$v = \int_{0}^{30} (6 - \sqrt{1.2} \cdot t^{1/2}) dt$$v = [6t - \sqrt{1.2} \cdot \frac{t^{3/2}}{3/2}]_{0}^{30}$$v = 6(30) - \frac{2}{3} \sqrt{1.2} \cdot (30)^{3/2}$$v = 180 - \frac{2}{3} \sqrt{1.2} \cdot 30 \sqrt{30} = 180 - 20 \sqrt{36} = 180 - 20(6) = 180 - 120 = 60 	ext{ ft/sec}$.Therefore, the maximum velocity is $v = 60 	ext{ ft/sec}$ at $T = 30 	ext{ sec}$.

The acceleration $f \text{ ft/sec}^2$ of a particle after a time $t \text{ sec}$ starting from rest is given by $f = 6 - \sqrt{1.2t}$. Then the maximum velocity $v$ and time $T$ to attain this velocity are:

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