Question

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

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

Answer 1

The Correct Option is 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, $\vec{r}_{0}=0$, $\vec{v}_{0}=5\hat{i}\,ms^{-1}$, $\vec{a}=\left(3\hat{i}+2\hat{j}\right)ms^{-2}$ $\therefore \vec{r}=5t\,\hat{i}+\frac{1}{2}\left(3\hat{i}+2\hat{j}\right)t^{2}$ $=\left(5t+1.5t^{2}\right)\hat{i}+1t^{2}\,\hat{j}\,...\left(i\right)$ 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 $t=6\,s$, $\vec{v}=23\hat{i}+12\hat{j}$ The speed of the particle is $\left|\vec{v}\right|=\sqrt{\left(23\right)^{2}+\left(12\right)^{2}}$ $ \approx 26\,ms^{-1}$