Question

The motion of a particle is defined by the position vector $\overrightarrow{r}=A(\cos t+t\sin t)\hat{i}+A(\sin t-t\cos t)\hat{j}$, where $t$ is expressed in seconds.

The position vector and acceleration vector are parallel

• A. at $t=1\text{s}$
• B. at $t=0$
• C. at $t=\sqrt{2}\text{s}$
• D. at $t=1.5\text{s}$

Answer 1

The correct option is B at $t=0$

We already got:
$\overrightarrow{r}=A(\cos t+t\sin t)\hat{i}+A(\sin t-t\cos t)\hat{j}$
$\overrightarrow{a}=A(\cos t-t\sin t)\hat{i}+A(\sin t+t\cos t)\hat{j}$

Now to find the time when both of these vectors are parallel, we can equate their cross product equal to zero.
$\overrightarrow{r}\times \overrightarrow{a}=\left[A^2\right(\cos t.\sin t+t{\cos }^{2}t+t{\sin }^{2}t+t^2\cos t.\sin t\left)\right]\hat{k}-\left[A^2\right(\sin t.\cos t-t{\cos }^{2}t-t{\sin }^{2}t+t^2\sin t\cos t\left)\right]\hat{k}$
$0=A^2(2t{\cos }^{2}t+2t{\sin }^{2}t)\hat{k}$
$0=A^2(2t\times 1)\hat{k}$
$\Rightarrow t=0s$