A particle's velocity changes from 3000 m/s in the y-direction to 4000 m/s in the x-direction over 20 seconds. Calculate its average acceleration.

The initial velocity is $3000 \text{ m/s}$ along the y-axis.

$$\vec{v_i} = 3000 \hat{j} \text{ m/s}$$

The final velocity is $4000 \text{ m/s}$ along the x-axis.

$$\vec{v_f} = 4000 \hat{i} \text{ m/s}$$

The change in velocity $\Delta \vec{v}$ is given by $\vec{v_f} - \vec{v_i}$.

$$\Delta \vec{v} = (4000 \hat{i} - 3000 \hat{j}) \text{ m/s}$$

The average acceleration $\vec{a}_{avg}$ is given by $\frac{\Delta \vec{v}}{\Delta t}$.
Given time interval $\Delta t = 20 \text{ s}$.

$$\vec{a}_{avg} = \frac{(4000 \hat{i} - 3000 \hat{j}) \text{ m/s}}{20 \text{ s}}$$ $$\vec{a}_{avg} = \left(\frac{4000}{20}\right) \hat{i} - \left(\frac{3000}{20}\right) \hat{j} \text{ m/s}^2$$ $$\vec{a}_{avg} = (200 \hat{i} - 150 \hat{j}) \text{ m/s}^2$$

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

$$|\vec{a}_{avg}| = \sqrt{(200 \text{ m/s}^2)^2 + (-150 \text{ m/s}^2)^2}$$ $$|\vec{a}_{avg}| = \sqrt{40000 + 22500} \text{ m/s}^2$$ $$|\vec{a}_{avg}| = \sqrt{62500} \text{ m/s}^2$$ $$|\vec{a}_{avg}| = 250 \text{ m/s}^2$$

The direction $\phi$ is given by $\tan\phi = \frac{a_y}{a_x}$.

$$\tan\phi = \frac{-150 \text{ m/s}^2}{200 \text{ m/s}^2}$$ $$\tan\phi = -0.75$$ $$\phi = \arctan(-0.75)$$ $$\phi \approx -36.87^\circ$$

Since the x-component is positive and the y-component is negative, the acceleration vector is in the fourth quadrant. The angle is $36.87^\circ$ below the positive x-axis.

The average acceleration has a magnitude of $\boxed{250 \text{ m/s}^2}$ and a direction of $\boxed{36.87^\circ \text{ below the positive x-axis}}$.