Question

The acceleration of a particle is given by $\overrightarrow{a}=\lfloor 2\hat{i}+6t\hat{j}+\frac{2{\pi }^2}{9}\cos \frac{\pi t}{3}\hat{k}\rfloor m/s^2$.
At $t=0,\overrightarrow{r}=0$ and $\overrightarrow{v}=(2\overrightarrow{i}+\hat{j})m{s}^1$. The position vector at $t=2s$ is

• A. $(8\hat{i}+10\hat{j}+\hat{k})m$
• B. $(8\hat{i}+10\hat{j}+3\hat{k})m$
• C. $(3\hat{i}+8\hat{j}+10\hat{k})m$
• D. $(10\hat{i}+3\hat{j}+8\hat{k})m$

Answer 1

The correct option is A $(8\hat{i}+10\hat{j}+\hat{k})m$

The acceleration of a particle is given by

$\overline{a}=[2\hat{i}+6t\hat{j}+\frac{2{\Pi }^2}{9}\cos \left(\frac{\Pi t}{3}\hat{k}\right)]$ $m/s^2$

at $t=0,\overline{r}=0$ and $\overline{v}=(2\hat{i}+\hat{j})m/s$

The acceleration= $\overline{a}=2\hat{i}+6t\hat{j}+\frac{2{\Pi }^2}{a}\cos \left[\frac{\Pi t\hat{i}}{3}\right]$

now, $t=29$

distance= $2\times 4\hat{i}+5\times 2\hat{j}+\frac{1}{2}\times \pi \hat{k}$

$=(8\hat{i}+10\hat{j}+\hat{k})m$

when $a=\frac{v}{t}$

$=\frac{(2\hat{i}+6t\hat{j}+\frac{2{\Pi }^2}{9}\cos \left(\frac{\Pi t}{3}\right)\hat{k}}{2}$

$\Rightarrow a=8\hat{i}+10\hat{j}+\hat{k}$ .