The average acceleration of an object is given by the formula:

$\vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t}$

where:

• $\Delta \vec{v}$ is the change in velocity,
• $\Delta t$ is the change in time.

Step 1: Determine the initial and final velocities

At $t = 1 \, \text{s}$:

• The velocity is $\vec{v}_1 = 3 \, \text{m/s}$ in the $+x$-direction. $\vec{v}_1 = 3 \hat{i} \, \text{m/s}$

At $t = 2 \, \text{s}$:

• The velocity is $\vec{v}_2 = 4 \, \text{m/s}$ in the $+y$-direction. $\vec{v}_2 = 4 \hat{j} \, \text{m/s}$

Step 2: Calculate the change in velocity

The change in velocity is:

$\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = (0 \hat{i} + 4 \hat{j}) - (3 \hat{i} + 0 \hat{j})$

$\Delta \vec{v} = -3 \hat{i} + 4 \hat{j} \, \text{m/s}$

Step 3: Calculate the average acceleration

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

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

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

This means the object is accelerating at $3 \, \text{m/s}^2$ in the negative $x$-direction and $4 \, \text{m/s}^2$ in the positive $y$-direction.

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Would you like more details, or do you have any questions?

Follow-up Questions:

1. How would you calculate the magnitude of the average acceleration?
2. What is the direction of the average acceleration in terms of angles?
3. Can the motion be described by any particular type of curve in the xy-plane?
4. How would the acceleration change if the velocities at different times were given?
5. How would you describe the motion in terms of kinematic equations?

Tip:

Remember that velocity and acceleration are vector quantities, so direction matters when calculating changes.