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(A) Given that the normal acceleration $a_c = \frac{v^2}{r}$ and tangential acceleration $a_t = \frac{dv}{dt}$ are equal: $\frac{v^2}{r} = \frac{dv}{d

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Given that the normal acceleration $a_c = \frac{v^2}{r}$ and tangential acceleration $a_t = \frac{dv}{dt}$ are equal: $\frac{v^2}{r} = \frac{dv}{dt}$ Integrating with respect to time from $t=0$ $(v=4 \ m/s)$ to $t$: $\int_{4}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{r}$ $\left[ -\frac{1}{v} \right]_{4}^{v} = \frac{t}{r}$ $-\frac{1}{v} + \frac{1}{4} = \frac{t}{0.5} = 2t$ $\frac{1}{v} = \frac{1}{4} - 2t = \frac{1-8t}{4} \implies v = \frac{4}{1-8t}$ Since $v = \frac{ds}{dt}$,we integrate to find the distance $s$ for one revolution $(s = 2\pi r = 2\pi(0.5) = \pi \ m)$: $\int_{0}^{\pi} ds = \int_{0}^{t} \frac{4}{1-8t} dt$ $\pi = 4 \left[ \frac{\ln(1-8t)}{-8} \right]_{0}^{t}$ $\pi = -\frac{1}{2} \ln(1-8t)$ $-2\pi = \ln(1-8t)$ $e^{-2\pi} = 1-8t$ $8t = 1 - e^{-2\pi}$ $t = \frac{1}{8} [1 - e^{-2\pi}] \ s$ Comparing with the given form,$\alpha = 8$.

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