A particle moves in 3D space with position vector $\mathbf{r}(t) = t^2 \hat{i} + (2t - 1) \hat{j} + t^3 \hat{k}$. The magnitude of the acceleration vector $|\mathbf{a}(t)|$ at time $t = 1$ is ________ (round off to two decimal places).
Answer Key:
6.32
Step-by-Step Solution & Explanation
Given the position vector:
$\mathbf{r}(t) = t^2 \hat{i} + (2t - 1) \hat{j} + t^3 \hat{k}$
Velocity is the first derivative:
$\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = 2t \hat{i} + 2 \hat{j} + 3t^2 \hat{k}$
Acceleration is the second derivative:
$\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = 2 \hat{i} + 0 \hat{j} + 6t \hat{k}$
At $t = 1$:
$\mathbf{a}(1) = 2 \hat{i} + 6 \hat{k}$
Magnitude of acceleration:
$|\mathbf{a}(1)| = \sqrt{2^2 + 0^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \approx 6.3246 \approx 6.32$