A particle moves along a space curve with position vector $\vec{r}(t) = (t^3)\hat{i} + (2t^2)\hat{j} + (3t)\hat{k}$. The magnitude of the particle's acceleration vector at time $t = 1$ is ________ (round off to 2 decimal places).
Answer Key:
7.21
Step-by-Step Solution & Explanation
Differentiating the position vector with respect to time $t$ to obtain velocity:
$\vec{v}(t) = \frac{d\vec{r}}{dt} = (3t^2)\hat{i} + (4t)\hat{j} + 3\hat{k}$
Differentiating velocity to find acceleration:
$\vec{a}(t) = \frac{d\vec{v}}{dt} = (6t)\hat{i} + 4\hat{j} + 0\hat{k}$
At $t = 1$:
$\vec{a}(1) = 6\hat{i} + 4\hat{j}$
The magnitude of acceleration is:
$|\vec{a}(1)| = \sqrt{6^2 + 4^2 + 0^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21$