Question

A particle travels so that its acceleration is given by $\overrightarrow{a}=5cost\hat{i}-3sint\hat{j}$. If the particle is located at $(-3,2)$ at time $t=0$ and is moving with a velocity given by $(-3\hat{i}+2\hat{j})$. Find
(i) the velocity $\left[\begin{array}{c}\overrightarrow{v}=\int \overrightarrow{a}.dt\end{array}\right]timet$
(ii) the position vector $[\overrightarrow{r}=\int \overrightarrow{v}.dt]$ of the particle at time $(t>0)$.

• A. (i)$\overrightarrow{v}=(5sint-3)\hat{i}+(3cost-1)\hat{j}$, (ii)$\overrightarrow{s}=(2-5cost-3t)\hat{i}+(2+3sint-t)\hat{j}$
• B. (i)$\overrightarrow{v}=(3sint-3)\hat{i}+(3cost-1)\hat{j}$, (ii)$\overrightarrow{s}=(2-5cost-3t)\hat{i}+(2+3sint-t)\hat{j}$
• C. (i)$\overrightarrow{v}=(4sint-3)\hat{i}+(3cost-1)\hat{j}$, (ii)$\overrightarrow{s}=(2-5cost-3t)\hat{i}+(2+3sint-t)\hat{j}$
• D. None of these

Answer 1

Answer: (A)

(i)$\overrightarrow{v}=(5sint-3)\hat{i}+(3cost-1)\hat{j}$, (ii)$\overrightarrow{s}=(2-5cost-3t)\hat{i}+(2+3sint-t)\hat{j}$

$\overrightarrow{a}=5cost\hat{i}-3sint\hat{j}$

$\Rightarrow \int d\overrightarrow{v}=\int 5costdt\hat{i}-\int 3sintdt\hat{j}$

Therefore $\underset{-3}{\stackrel{v}{\int }}d{v}_x=\underset{0}{\stackrel{t}{\int }}5costdt\Rightarrow v_x=5sint-3$

$\frac{dx}{dt}=(5sint-3)\Rightarrow \underset{-3}{\stackrel{x}{\int }}dx=\underset{0}{\stackrel{t}{\int }}(5sint-3)dt$

$x+3=5-5cost-3t\Rightarrow x=2-5cost-3t$

Similarly,

$\underset{2}{\stackrel{v}{\int }}d{v}_y=-\underset{0}{\stackrel{t}{\int }}3sintdt$

$\Rightarrow v_y-2=3(cost-1)\Rightarrow v_y=3cost-1$

$\Rightarrow \underset{2}{\stackrel{y}{\int }}dy=\underset{0}{\stackrel{t}{\int }}(3cost-1)dt$

$\Rightarrow y-2=3sint-t\Rightarrow y=2+3sint-t$

Thus, $\overrightarrow{v}=(5sint-3)\hat{i}+(3cost-1)\hat{j}$

and $\overrightarrow{s}=(2-5cost-3t)\hat{i}+(2+3sint-t)\hat{j}$