an object is moving in the xy plane. At t=1 seconds, the object's velocity is 3 m/s in the +x direction. At t=2 seconds, the object's velocity is 4m/s in the +y direction. What is the average accleration of the object between t=1 second and t=2 seconds? Please explain

To calculate the average acceleration of the object between \( t = 1 \) second and \( t = 2 \) seconds, we need to use the formula for average acceleration:

\[
\mathbf{a_{avg}} = \frac{\Delta \mathbf{v}}{\Delta t}
\]

Where:
- \( \mathbf{a_{avg}} \) is the average acceleration (a vector quantity),
- \( \Delta \mathbf{v} \) is the change in velocity (a vector),
- \( \Delta t \) is the time interval over which the change occurs.

### Step 1: Initial and Final Velocity Vectors
- At \( t = 1 \) second, the velocity of the object is \( \mathbf{v_1} = 3 \, \text{m/s} \) in the \( +x \) direction. This can be written as:
  \[
  \mathbf{v_1} = (3 \, \hat{i}, 0 \, \hat{j}) \, \text{m/s}
  \]
  where \( \hat{i} \) is the unit vector in the \( x \)-direction and \( \hat{j} \) is the unit vector in the \( y \)-direction.

- At \( t = 2 \) seconds, the velocity of the object is \( \mathbf{v_2} = 4 \, \text{m/s} \) in the \( +y \) direction. This can be written as:
  \[
  \mathbf{v_2} = (0 \, \hat{i}, 4 \, \hat{j}) \, \text{m/s}
  \]

### Step 2: Calculate the Change in Velocity
The change in velocity \( \Delta \mathbf{v} \) is the difference between the final velocity \( \mathbf{v_2} \) and the initial velocity \( \mathbf{v_1} \):
\[
\Delta \mathbf{v} = \mathbf{v_2} - \mathbf{v_1} = (0 \, \hat{i}, 4 \, \hat{j}) - (3 \, \hat{i}, 0 \, \hat{j})
\]
\[
\Delta \mathbf{v} = (-3 \, \hat{i}, 4 \, \hat{j}) \, \text{m/s}
\]

### Step 3: Time Interval
The time interval \( \Delta t \) is the difference between the two time instants, which is:
\[
\Delta t = 2 \, \text{seconds} - 1 \, \text{second} = 1 \, \text{second}
\]

### Step 4: Calculate the Average Acceleration
Using the formula for average acceleration:
\[
\mathbf{a_{avg}} = \frac{\Delta \mathbf{v}}{\Delta t} = \frac{(-3 \, \hat{i}, 4 \, \hat{j})}{1 \, \text{second}}
\]
\[
\mathbf{a_{avg}} = (-3 \, \hat{i}, 4 \, \hat{j}) \, \text{m/s}^2
\]

So, the average acceleration vector is:
\[
\mathbf{a_{avg}} = -3 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}^2
\]

### Step 5: Magnitude of the Average Acceleration
To find the magnitude of the average acceleration, we use the Pythagorean theorem:
\[
|\mathbf{a_{avg}}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2
\]

### Final Answer:
- The average acceleration vector is \( -3 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}^2 \).
- The magnitude of the average acceleration is \( 5 \, \text{m/s}^2 \).

Would you like more details on any step or have any questions?

### Related Questions:
1. How do you calculate the direction of the average acceleration from its components?
2. What would happen if the velocities were given in different directions at the start and end?
3. How does the average acceleration differ from instantaneous acceleration?
4. Can we apply the same process if the object was moving in three dimensions?
5. What if the time interval between measurements was not exactly 1 second?

### Tip:
When dealing with vector quantities like velocity and acceleration, always break them into components and handle each direction separately to avoid confusion.

An object is moving in the xy plane. At t=1 second, the object's velocity is 3 m/s in the +x direction. At t=2 seconds, the object's velocity is 4 m/s in the +y direction. What is the average acceleration of the object between t=1 second and t=2 seconds? Please explain.

Learn how to calculate the average acceleration of an object moving in the xy plane between t=1 second and t=2 seconds. This solution covers kinematic equations, vector decomposition, and the use of the Pythagorean theorem to find the magnitude of acceleration. Suitable for students in Grades 9-12.

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