Question

A particle leaves the origin with an initial velocity $\overrightarrow{u}=3\hat{i}m/s$ and a constant acceleration $\overrightarrow{a}=(-1.0\hat{i}-0.5\hat{j})$. its velocity $\overrightarrow{v}$ and position vector $\overrightarrow{r}$ when it reaches its maximum x-co-ordinate are:-

• A. $\overrightarrow{v}=-2\hat{j}$
• B. $\overrightarrow{v}=1.5\hat{j}m/s$
• C. $\overrightarrow{r}=(4.5\hat{i}+2.25\hat{j})m$
• D. $\overrightarrow{r}=(3\hat{i}+2\hat{j})m$

Answer 1

The correct option is C $\overrightarrow{r}=(4.5\hat{i}+2.25\hat{j})m$

Given,

$\overrightarrow{u}=3\hat{i}m/sor3m/sinx-direction$

$\overrightarrow{a}=(-1.0\hat{i}-0.5\hat{j})or-1m/s$ in x-direction and $-0.5m/s$ in y-direction.

When it reaches its maximum x-coordinate, its velocity in x-direction will be zero. Let $t$ be that time, then

$v_x=u_x+a_{x}t$

$\Rightarrow 0=3-1\left(t\right)(\because v_x=0)$

$\Rightarrow t=3sec$

$v_y=u_y+a_{y}t$

$\Rightarrow v_y=0+(-0.5)\left(3\right)$

$\Rightarrow v_y=-1.5m/s$

$\Rightarrow \overrightarrow{v}=-1.5\hat{j}m/s$

Using $3^{rd}$ equation of motion

$v_x^2=u_x^2+2a_x{s}_x$

$\Rightarrow s_x=\frac{(v_x^2-u_x^2)}{2a_x}$

$\Rightarrow s_x=\frac{(0^2-3^2)}{2(-1)}=4.5m$

Similarly,

$\Rightarrow s_y=\frac{(v_y^2-u_y^2)}{2a_y}$

$\Rightarrow s_x=\frac{(-1.5{)}^2-0^2}{2(-0.5)}=2.25m$

So, $\overrightarrow{r}=(4.5\hat{i}+2.25\hat{j})m/s$

So, the correct option will be $\left(C\right)$