Question

A particle leaves the origin with an initial velocity $\vec{u}=(3\hat{i})m{s}^{-1}$ and at a constant acceleration $\vec{a}=(-1.0\hat{i}-0.5\hat{j})m{s}^{-2}$. Its velocity $v$ and position vector $\vec{r}$ when it reaches its maximum $x$-coordinate are
(1) $v=-2\hat{j}m{s}^{-1}$
(2) $v=(-1.5\hat{j})m{s}^{-1}$
(3) $\vec{r}=(4.5\hat{i}-1.25\hat{j})m$
(4) $\vec{r}=(4.5\hat{i}-2.25\hat{j})m$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (C)

$u_x=3m{s}^{-1},a_x=-1.0m{s}^{-2}$
$\therefore v_x^2=u_x^2+2a_{x}x$
or $0=(3{)}^2+2(-1)(x)$
or $x=4.5m$
Also $v_x=u_x+a_{x}t$
$0=3-1.0t$
or $t=3s$
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2=0+\frac{1}{2}(-0.5)(3{)}^2=-2.25m$
and $v_y=a_{y}t=(-0.5)(3)=-1.5m{s}^{-1}$
$\therefore v=v_x\hat{i}+v_y\hat{j}=0-1.5\hat{j}=(-1.5\hat{j})m{s}^{-1}$
and $\vec{r}=x\hat{i}+\hat{y}=(4.5\hat{i}-2.25\hat{j})m$