Question

A particle moves such that its position vector is given by $\vec r = \cos(3t)\,\mathbf{i} + \sin(3t)\,\mathbf{j}$. Which of the following is/are incorrect?

• A. Velocity is parallel to $\vec r$
• B. Velocity is antiparallel to $\vec r$
• C. Acceleration is antiparallel to $\vec r$

Answer 1

Answer: (A), (B)

Velocity is parallel to $\vec r$; Velocity is antiparallel to $\vec r$

Answer 2

Solution Approach:

1. Position vector:

$$\vec r = \cos(3t)\,\mathbf{i} + \sin(3t)\,\mathbf{j}$$

2. Velocity:

$$\vec v = \frac{d\vec r}{dt} = -3\sin(3t)\,\mathbf{i} + 3\cos(3t)\,\mathbf{j}$$

Notice that $\vec v\cdot\vec r = \bigl(-3\sin(3t)\bigr)\cos(3t) + \bigl(3\cos(3t)\bigr)\sin(3t) = 0$.
Hence, $\vec v$ is perpendicular to $\vec r$, so it is neither parallel nor antiparallel.

3. Acceleration:

$$\vec a = \frac{d\vec v}{dt} = -9\cos(3t)\,\mathbf{i} - 9\sin(3t)\,\mathbf{j} = -9\,\vec r$$

This shows $\vec a$ is antiparallel to $\vec r$.

Therefore, the incorrect statements are that velocity is parallel or antiparallel to $\vec r$.