Question

A particle starts from origin at $t=0$ with a velocity $5\hat{i}m{s}^{-1}$ and moves in $x$-$y$ plane under the action of a force which produces a constant acceleration of $3\hat{i}+2\hat{j}m{s}^{-2}$, the speed of the particle at this time is

• A. $16m{s}^{-1}$
• B. $26m{s}^{-1}$
• C. $36m{s}^{-1}$
• D. $46m{s}^{-1}$

Answer 1

Answer: (B)

The position of the particle is given by
$\vec{r}={\vec{r}}_0+{\vec{v}}_{0}t+\frac{1}{2}\vec{a}{t}^2$
where, ${\vec{r}}_0$ is the position vector at $t=0$ and ${\vec{v}}_0$ is the velocity at $t=0$
Here, ${\vec{r}}_0=0$,
${\vec{v}}_0=5\hat{i}m{s}^{-1}$,
$\vec{a}=\left(3\hat{i}+2\hat{j}\right)m{s}^{-2}$
$\therefore \vec{r}=5t\hat{i}+\frac{1}{2}\left(3\hat{i}+2\hat{j}\right)t^2$
$=\left(5t+1.5t^2\right)\hat{i}+1t^2\hat{j}...\left(i\right)$
Velocity,
$\vec{v}=\frac{d\vec{r}}{dt}=\frac{d}{dt}\left(5t+1.5t^2\right)\hat{i}+1t^2\hat{j}$
$=\left(5+3t\right)\hat{i}+2t\hat{j}$
At $t=6s$,
$\vec{v}=23\hat{i}+12\hat{j}$
The speed of the particle is
$\left|\vec{v}\right|=\sqrt{{\left(23\right)}^2+{\left(12\right)}^2}$
$\approx 26m{s}^{-1}$