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

Answer: (A)

Step 1: Given that:

The position vector of the particle in motion is given as;

$\overrightarrow{r}=A(\cos t+t\sin t)\hat{i}+A(\sin t-t\cos t)\hat{j}$

Step 2: Concept and formula used:

The velocity of a body in differential form is given by;

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

And the acceleration in the body in differential form is given by;

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$

1. ​Dot product of two vectors,$\overrightarrow{a}$ and $\overrightarrow{b}$ is defined as, $\overrightarrow{a}.\overrightarrow{b}=ab\cos \theta$ , where $\theta$ is the angle between both vectors.
2. Two vectors are said to be perpendicular to each other, if the dot product of both the vectors is equal to zero.
3. In the case of dot product; if we have two vectors, such as

$\overrightarrow{a}=a_x\hat{i}+a_y\hat{j}$

$\overrightarrow{b}=b_x\hat{i}+b_y\hat{j}$

Then the dot product is defined as,

$\overrightarrow{a}.\overrightarrow{b}=(a_x\hat{i}+a_y\hat{j}).(b_x\hat{i}+b_y\hat{j})$

$\overrightarrow{a}.\overrightarrow{b}=a_x{b}_x+a_y{b}_y$

As, $\hat{i}.\hat{i}=\hat{j}.\hat{j}=1,but,\hat{i}.\hat{j}=\hat{j}.\hat{i}=0$

Thus,

For being perpendicular, the dot product of the position vector and acceleration vector must be perpendicular.

Step 3: Calculation of the acceleration in the particle:

$\overrightarrow{v}=\frac{d}{dt}(A(\cos t+t\sin t)\hat{i}+A(\sin t-t\cos t)\hat{j})$

$\overrightarrow{v}=\frac{d}{dt}(A\cos t\hat{i}+At\sin t\hat{i}+A\sin t\widehat{j-At\cos t\hat{j}})$

$\overrightarrow{v}=-A\sin t\widehat{i+}\left(A\sin t\widehat{\hat{i}+At\cos t\hat{i}}\right)+A\cos t\hat{j}-\left(A\cos t\widehat{\hat{j}-At\sin t\hat{j}}\right)$

$\overrightarrow{v}=A(-\sin t+\sin t+t\cos t)\hat{i}+A(\cos t-\cos t+t\sin t)\hat{j}$

$\overrightarrow{v}=At\cos t\hat{i}+At\sin t\hat{j}$

$\overrightarrow{v}=At(\cos t\hat{i}+\sin t\hat{j})$

Now, the acceleration of the particle will be;

$\overrightarrow{a}=\frac{d}{dt}\overrightarrow{v}$

$\overrightarrow{a}=At(\cos t\hat{i}+\sin t\hat{j})$

$\overrightarrow{a}=A(t\cos t\hat{i}+t\sin t\hat{j})$

$\overrightarrow{a}=A(\cos t\hat{i}-t\sin t\hat{i})+A(\sin t\hat{j}+t\cos t\hat{j})$

$\overrightarrow{a}=A(\cos t-t\sin t)\hat{i}+A(\sin t+t\cos t)\hat{j}$

Step 4: Determining the condition for the position vector and acceleration vector to be perpendicular:

Now,

$\overrightarrow{r}.\overrightarrow{a}=0$

$\{A(\cos t+t\sin t)\hat{i}+A(\sin t-t\cos t)\hat{j}\}.\{A(\cos t-t\sin t)\hat{i}+A(\sin t+t\cos t)\hat{j}\}=0$

$A^2(\cos t+t\sin t)(\cos t-t\sin t)+A^2(\sin t-t\cos t)(\sin t+t\cos t)=0$

$A^2\{{\cos }^{2}t-t^2{\sin }^{2}t\}+A^2\{{\sin }^{2}t-t^2{\cos }^{2}t\}=0$

$A^2\{{\cos }^{2}t+{\sin }^{2}t-t^2({\sin }^{2}t+{\cos }^{2}t)\}=0$

$A^2\times (1-t^2\times 1)=0$

$(\because {\cos }^{2}t+{\sin }^{2}t={\sin }^{2}t+{\cos }^{2}t=1)$

$A^2(1-t^2)=0$

Since,

$A^2\ne 0,$

$(1-t^2)=0$

$t^2=1$

$t=1sec$

Thus,

Option C) at $t=1sec$ is the correct option.