Question

If the position of the particle as a function of time \( t \) is \( \vec{r} = 8t \hat{i} + 3t^2 \hat{j} + 3 \hat{k} \, \text{m} \), then the acceleration of the particle is (in \( \text{m/s}^2 \))

• A. 6
• B. 3
• C. 8
• D. 4
• E. 5

Hint:

To find acceleration from position, first differentiate position to get velocity, then differentiate velocity to get acceleration.

Answer 1

The Correct Option is A

The position of the particle is given by: \[ \vec{r}(t) = 8t \hat{i} + 3t^2 \hat{j} + 3 \hat{k} \] To find the acceleration, we first need to find the velocity and then differentiate it to get the acceleration. 1. The velocity \( \vec{v}(t) \) is the first derivative of the position with respect to time: \[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(8t \hat{i} + 3t^2 \hat{j} + 3 \hat{k}) \] \[ \vec{v}(t) = 8 \hat{i} + 6t \hat{j} + 0 \hat{k} \] 2. The acceleration \( \vec{a}(t) \) is the first derivative of velocity with respect to time: \[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(8 \hat{i} + 6t \hat{j} + 0 \hat{k}) \] \[ \vec{a}(t) = 0 \hat{i} + 6 \hat{j} + 0 \hat{k} \] Thus, the acceleration is: \[ \vec{a}(t) = 6 \hat{j} \, \text{m/s}^2 \] The magnitude of the acceleration is \( 6 \, \text{m/s}^2 \). Thus, the correct answer is (A) 6.