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Question

(A) Given the position vector: $\vec{r} = 3t^2 \hat{i} + 2t \hat{j} + \hat{k}$.Velocity is the time derivative of position: $\vec{v} = \frac{d\vec{r}}

Vedclass · Accepted Answer

$6, \sqrt{148}$

Vedclass · Answer

(A) Given the position vector: $\vec{r} = 3t^2 \hat{i} + 2t \hat{j} + \hat{k}$.Velocity is the time derivative of position: $\vec{v} = \frac{d\vec{r}}{dt} = 6t \hat{i} + 2 \hat{j}$.At $t = 2 	ext{ s}$, the velocity vector is $\vec{v} = 6(2) \hat{i} + 2 \hat{j} = 12 \hat{i} + 2 \hat{j}$.The magnitude of velocity is $|\vec{v}| = \sqrt{12^2 + 2^2} = \sqrt{144 + 4} = \sqrt{148} 	ext{ m/s}$.Acceleration is the time derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt} = 6 \hat{i}$.The magnitude of acceleration is $|\vec{a}| = \sqrt{6^2} = 6 	ext{ m/s}^2$.Thus, the magnitude of acceleration is $6$ and the magnitude of velocity is $\sqrt{148}$.

The magnitude of acceleration and velocity of a particle moving in a plane, whose position vector is $\vec{r} = 3t^2 \hat{i} + 2t \hat{j} + \hat{k}$ at $t = 2 \text{ s}$, are, respectively:

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