Find the magnitude of velocity and acceleration at t=2 if r = (4t2 − t3)i − 5t j + (t2 - 2t)k

Velocity is the first derivative of the position vector with respect to time.

$$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(4t^2 - t^3)\hat{i} - \frac{d}{dt}(5t)\hat{j} + \frac{d}{dt}(t^4 - 2t)\hat{k}$$ $$\vec{v}(t) = (8t - 3t^2)\hat{i} - 5\hat{j} + (4t^3 - 2)\hat{k}$$

Substitute $t=2$ into the velocity vector equation:

$$\vec{v}(2) = (8(2) - 3(2)^2)\hat{i} - 5\hat{j} + (4(2)^3 - 2)\hat{k}$$ $$\vec{v}(2) = (16 - 3(4))\hat{i} - 5\hat{j} + (4(8) - 2)\hat{k}$$ $$\vec{v}(2) = (16 - 12)\hat{i} - 5\hat{j} + (32 - 2)\hat{k}$$ $$\vec{v}(2) = 4\hat{i} - 5\hat{j} + 30\hat{k}$$

The magnitude of a vector $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ is $|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$.

$$|\vec{v}(2)| = \sqrt{4^2 + (-5)^2 + 30^2}$$ $$|\vec{v}(2)| = \sqrt{16 + 25 + 900}$$ $$|\vec{v}(2)| = \sqrt{941}$$

The magnitude of velocity at $t=2$ is $\boxed{\sqrt{941}}$.

Acceleration is the first derivative of the velocity vector with respect to time.

$$\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(8t - 3t^2)\hat{i} - \frac{d}{dt}(5)\hat{j} + \frac{d}{dt}(4t^3 - 2)\hat{k}$$ $$\vec{a}(t) = (8 - 6t)\hat{i} - 0\hat{j} + (12t^2)\hat{k}$$ $$\vec{a}(t) = (8 - 6t)\hat{i} + 12t^2\hat{k}$$

Substitute $t=2$ into the acceleration vector equation:

$$\vec{a}(2) = (8 - 6(2))\hat{i} + 12(2)^2\hat{k}$$ $$\vec{a}(2) = (8 - 12)\hat{i} + 12(4)\hat{k}$$ $$\vec{a}(2) = -4\hat{i} + 48\hat{k}$$

$$|\vec{a}(2)| = \sqrt{(-4)^2 + 0^2 + 48^2}$$ $$|\vec{a}(2)| = \sqrt{16 + 0 + 2304}$$ $$|\vec{a}(2)| = \sqrt{2320}$$

The magnitude of acceleration at $t=2$ is $\boxed{\sqrt{2320}}$.