To find the average acceleration of the object, we need to use the formula for average acceleration:

$\vec{a}_{avg} = \frac{\vec{v}_2 - \vec{v}_1}{t_2 - t_1}$

Where:

• $\vec{a}_{avg}$ is the average acceleration vector,
• $\vec{v}_1$ and $\vec{v}_2$ are the velocity vectors at times $t_1$ and $t_2$,
• $t_1 = 1 \, \text{s}$,
• $t_2 = 2 \, \text{s}$,
• $\vec{v}_1 = 3 \, \text{m/s} \, \hat{i}$ (velocity in the +x direction),
• $\vec{v}_2 = 4 \, \text{m/s} \, \hat{j}$ (velocity in the +y direction).

Now let's break down the change in velocity:

1. The velocity at $t_1 = 1 \, \text{s}$ is $\vec{v}_1 = 3 \, \hat{i} + 0 \, \hat{j} \, \text{m/s}$.
2. The velocity at $t_2 = 2 \, \text{s}$ is $\vec{v}_2 = 0 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}$.

The change in velocity is: $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = (0 \, \hat{i} + 4 \, \hat{j}) - (3 \, \hat{i} + 0 \, \hat{j})$ $\Delta \vec{v} = -3 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}$

The time interval is: $\Delta t = t_2 - t_1 = 2 \, \text{s} - 1 \, \text{s} = 1 \, \text{s}$

Now, the average acceleration is: $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{-3 \, \hat{i} + 4 \, \hat{j}}{1 \, \text{s}} = -3 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}^2$

Thus, the average acceleration is: $\vec{a}_{avg} = -3 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}^2$

In magnitude, the average acceleration can be found as: $|\vec{a}_{avg}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2$

So the magnitude of the average acceleration is $5 \, \text{m/s}^2$, and the direction is in the quadrant where the acceleration is negative in the $x$-direction and positive in the $y$-direction.

Would you like a detailed breakdown of any part of this solution?

Here are 5 related questions:

1. How is the direction of acceleration determined in 2D motion?
2. What happens if the velocity changes both in magnitude and direction?
3. How does the average acceleration differ from instantaneous acceleration?
4. How do we calculate the magnitude of a vector from its components?
5. How would the result change if the velocity at $t=2s$ had components in both $x$ and $y$?

Tip: The average acceleration only depends on the initial and final velocities, not on the path the object takes in between.