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 perpendicular

• 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 A at $t=1\text{s}$

We can write for x-direction,
$r_x=A(\cos t+t\sin t)$
$v_x=\frac{d{r}_x}{dt}=-A\sin t+A\sin t+At\cos t$
$\Rightarrow v_x=At\cos t$
$a_x=\frac{d{v}_x}{dt}=A(\cos t-t\sin t)$

Also we can solve for y-direction:
$r_y=A(\sin t-t\cos t)$
$v_y=A\left(t\sin t\right)$
$a_y=A(\sin t+t\cos t)$

To find out time when the position vector and acceleration vector are perpendicular, we can equate their dot product to zero.

$\overrightarrow{r}.\overrightarrow{a}=r_x\times a_x+r_y\times a_y=0$
$0=A({\cos }^{2}t-t^2{\sin }^{2}t)+A({\sin }^{2}t-t^2{\cos }^{2}t)$
$0=A[{\cos }^{2}t+{\sin }^{2}t-t^2({\sin }^{2}t+{\cos }^{2}t\left)\right]$
$0=A[1-t^2(1\left)\right]$
$\Rightarrow t=1s$