Question

A ponit $P$ is moving in uniform circular motion with radius $3\, m$. Let at some instant the acceleration of the point is $\vec{a}=(6 \hat{i}-4 \hat{j}) \,m / s^{2}$, the position vector is $\bar{r}$ and velocity is $\vec{r}$ and velocity vector is $\vec{v}$. The correct statement is

• A. $\vec{v} \cdot \vec{a}=0$ and $\vec{r} \times \vec{a} \neq 0$
• B. $\vec{v} \cdot \vec{a} \neq 0$ and $\vec{r} \times \vec{a} \neq 0$
• C. $\vec{v} \cdot \vec{a}=0$ and $\vec{r} \times \vec{a}=0$
• D. $\vec{v} \cdot \vec{a} \neq 0$ and $\vec{r} \times \vec{a}=0$

Answer 1

Answer: (C)

The given situation is shown in the following figure

Acceleration, $a=6 \hat{i}-4 \hat{j}$
Clearly, acceleration $a$ is perpendicular to velocity $v$, hence $v \cdot a=0$
Again, position vector $r$ is parallel to acceleration $a$, hence $r \times a=0$