A particle is moving in a circle of radius 50cm in such a way that...

Given:

$\left|\overset{\to}{a}c\right| = \left|\overset{\to}{a}t\right|$

$⇒ \dfrac{{v}^{2}}{r} = \dfrac{dv}{dt}$

$⇒ {∫}_{4}^{v}\dfrac{dv}{{v}^{2}} = {∫}_{0}^{t}\dfrac{dt}{r}$

$⇒ {\left[\dfrac{- 1}{v}\right]}_{4}^{v} = \dfrac{t}{r}$

$⇒ \dfrac{- 1}{v} + \dfrac{1}{4} = 2t$

$⇒ v = \dfrac{4}{1 - 8t} = \dfrac{ds}{dt}$

$⇒ 4{∫}_{0}^{t}\dfrac{dt}{1 - 8t} = {∫}_{0}^{s}ds$

\left(r = 0 . 5m, \text{distance}\text{covered}s = 2\pir = \pi\right)$</p><p>$⇒ 4 \times \dfrac{[ \ln ( 1 - 8t ) {]}_{0}^{t}}{- 8} = \pi$</p><p>$⇒ ℓn ( 1 - 8t ) = - 2\pi$</p><p>$⇒ 1 - 8t = {e}^{- 2\pi}$</p><p>$⇒ t = \left(1 - {e}^{- 2\pi}\right)\dfrac{1}{8}s$</p><p>So,$\alpha = 8$.