At time t,the force acting on a particle P of mass 2kg is (2ti + 4j)N.P is initially at rest at the point with position vector (i + 2j). Find: a) the velocity of P when t = 2. b) the position vector when t = 2.


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Question Number 38366 by Rio Mike last updated on 24/Jun/18

$A t t i m e t, t h e f o r c e a c t i n g o n a p a r t i c l e$ $P o f m a s s 2 k g i s (2 t i + 4 j) N . P$ $i s i n i t i a l l y a t r e s t a t t h e p o i n t w i t h$ $p o s i t i o n v e c t o r (i + 2 j) .$ $F i n d :$ $a) t h e v e l o c i t y o f P w h e n t = 2 .$ $b) t h e p o s i t i o n v e c t o r w h e n t = 2 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

	$m \frac{d \vec{v}}{d t} = 2 t \vec{i} + 4 \vec{j}$ $\frac{d \vec{v}}{d t} = t \vec{i} + 2 \vec{j} m = 2 k g$ $d \vec{v} = t \vec{i d t} + 2 \vec{j d t}$ $\vec{v} = \frac{t^{2}}{2} \vec{i} + 2 \vec{j t} + c$ $t = 0 \vec{v} = 0 c = 0$ $\vec{v} = \frac{t^{2}}{2} \vec{i} + 2 t \vec{j}$ $\vec{v} = 2 \vec{i} + 4 \vec{j} w h e n t = 2$ $\frac{d \vec{r}}{d t} = 2 \vec{i} + 4 \vec{j}$ $\overset{}{r} = 2 t \vec{i} + 4 t \overset{\to r}{j} + c$ $w h e n t = 0 {\vec{r}}_{0} = \vec{i} + 2 \vec{j}$ $\vec{r} = 2 t \vec{i} + 4 t \vec{j} + \vec{i} + 2 \vec{j}$ $\vec{r} = \vec{i} (2 t + 1) + \vec{j} (4 t + 2)$
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