4kg ball falls from rest at time t =0 in a medium offering a resistance in kg numerically equal to twice its instantaneous velocity in m/s. find; (a) the velocity and distance travelled at any time t>0 (b) the limiting velocity


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Question Number 28375 by Ruchinna1 last updated on 25/Jan/18

$4 k g b a l l f a l l s f r o m r e s t a t t i m e t = 0 i n a$ $m e d i u m o f f e r i n g a r e s i s t a n c e i n k g$ $n u m e r i c a l l y e q u a l t o t w i c e i t s i n s t a n t a n e o u s$ $v e l o c i t y i n m / s .$ $f i n d;$ $(a) t h e v e l o c i t y a n d d i s t a n c e t r a v e l l e d a t a n y$ $t i m e t > 0$ $(b) t h e l i m i t i n g v e l o c i t y$
	Answered by mrW2 last updated on 25/Jan/18

	$R = 2 v [k g] = 2 g v [N]$ $m g - R = m a = m \frac{d v}{d t}$ $m g - 2 g v = m \frac{d v}{d t}$ $4 g - 2 g v = 4 \frac{d v}{d t}$ $\Rightarrow 5 (2 - v) = \frac{d v}{d t}$ $\Rightarrow \frac{d v}{2 - v} = 5 d t$ $\Rightarrow \int_{0}^{v} \frac{d v}{2 - v} = 5 \int_{0}^{t} d t$ $\Rightarrow {[- \ln (2 - v)]}_{0}^{v} = 5 {[t]}_{0}^{t}$ $- \ln \frac{2 - v}{2} = 5 t$ $1 - \frac{v}{2} = e^{- 5 t}$ $\Rightarrow v = 2 (1 - e^{- 5 t})$ $\Rightarrow \frac{d s}{d t} = 2 (1 - e^{- 5 t})$ $\Rightarrow s = 2 \int_{0}^{t} (1 - e^{- 5 t}) d t = 2 (t + \frac{e^{- 5 t} - 1}{5}) = \frac{2}{5} (5 t + e^{- 5 t} - 1)$ $\lim_{t \to \infty} v = 2 m / s$
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