Question

A particle leaves the origin with an initial velocity $\overrightarrow{v}$ = ( $3.00\hat{i}$ ) m/s and a constant acceleration $\overrightarrow{a}$ = ( $-1.00\hat{i}-0.500\hat{j}$ ) m/$s^2$. When it reaches its maximum x coordinate, match its position vector and velocity vector(Coloumn A) with vectors (Coloumn B)?

Coloumn A	Coloumn B
(i) Position vector	(U) $-10\hat{j}$
(ii) Velocity vector	(V) $4.5\hat{i}-22.5\hat{j}$
	(W) $4.5\hat{i}-21.5\hat{j}$
	(X) $-15\hat{j}$

• A. (i) - U, (ii) - V
• B. (i) - X, (ii) - W
• C. (i) - V, (ii) - X
• D. (i) - W, (ii) - X

Answer 1

The correct option is C

(i) - V, (ii) - X

$\overrightarrow{u}$ = $3\hat{i}$ m/s , $\overrightarrow{a}$ = ( $-1\hat{i}-5\hat{j}$ ) m/$s^2$

This can be treated as 2 motions - one in x direction and the other in y direction.

x direction

$s_x=ut+\frac{1}{2}a{t}^2$ ( Let $s_x$ be the displacement in x direction )

u = 3m/s , a = -1m/$s^2$ ( x component of inital velocity is 3m/s and x component ot $\overrightarrow{a}$ is -1m/$s^2$ )

$\therefore$ $s_x=3t+\frac{1}{2}(-1)t^2$

$\Rightarrow$ $s_x=3t-\frac{1}{2}{t}^2$

$v_x=u+at$

= 3 + (-1)t

=(3-t)s

x component will be maximum , when $\frac{d{s}_x}{dt}$ or $v_x$ is zero

If we draw a graph between $s_x$ and t , we get

We can see that $s_x$ is maximum at t = 3 or when $\frac{d{s}_x}{dt}$ = 0 or $v_x$ = 0

Similarly in y direction

u = 0 ( as y component of inital velocity is zero )

a = -5m/$s^2$ ( y component of acceleration )

$s_y=ut+\frac{1}{2}a{t}^2$

= 0 * (t) + $\frac{1}{2}(-5)t^2$

= $\frac{-5}{2}{t}^{2}m$

And $v_y$ or $\frac{d{s}_y}{dt}$ = u + at

= 0 + (-5)t

= -5t m/s

Now at t = 3s

$s_x$ = 3t- $\frac{1}{2}{t}^2$ = 3(3) - $\frac{1}{2}$ (3)^2 = 9 - $\frac{9}{2}$ = 4.5m

$s_y$ = $\frac{-5}{2}(t{)}^2=\frac{-5}{2}\times 9=-\frac{45}{3}=-22.5m$

$v_x$ = 3 - t = 3 - 3 = 0 m/s

$v_y$ = -5t = -5(3) = -15m/s

$\therefore$ Position vector = (4.5$\hat{i}$ - 22.5$\hat{j}$)m

Velocity vector = -15$\hat{j}$ m/s