https://file.aiquickdraw.com/mathbot/file/17266957787477lozkr6u.jpg

The figure shows the force acting on a 1.0 kg object as it moves along the x-axis. The object is at rest at the origin at t = 0 s. What are its acceleration and velocity at t = 6 s?

To solve for the **acceleration** and **velocity** of the object at \( t = 6 \, \text{s} \), let's go through the steps systematically using the given force-time graph:

### 1. Mass of the Object
The mass of the object is given as \( m = 1.0 \, \text{kg} \).

### 2. Newton’s Second Law of Motion
We know from Newton's second law that:
\[
F = m a
\]
This can be rewritten as:
\[
a = \frac{F}{m}
\]
Thus, for each interval, we will find the acceleration by dividing the force by the mass.

### 3. Calculating Acceleration in Each Time Interval
The force is given as a step function in the graph, so the force acting on the object varies with time. The time intervals and corresponding forces are:
- From \( t = 0 \, \text{s} \) to \( t = 2 \, \text{s} \), \( F_x = 4 \, \text{N} \)
- From \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \), \( F_x = 0 \, \text{N} \)
- From \( t = 4 \, \text{s} \) to \( t = 6 \, \text{s} \), \( F_x = -2 \, \text{N} \)

Using \( a = \frac{F}{m} \), let's compute the acceleration in each interval:
- For \( t = 0 \, \text{s} \) to \( t = 2 \, \text{s} \), \( a = \frac{4}{1.0} = 4 \, \text{m/s}^2 \)
- For \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \), \( a = \frac{0}{1.0} = 0 \, \text{m/s}^2 \)
- For \( t = 4 \, \text{s} \) to \( t = 6 \, \text{s} \), \( a = \frac{-2}{1.0} = -2 \, \text{m/s}^2 \)

### 4. Velocity Calculations
The object is initially at rest at \( t = 0 \, \text{s} \), so \( v_0 = 0 \, \text{m/s} \). Now, we can calculate the velocity after each time interval.

- **From \( t = 0 \, \text{s} \) to \( t = 2 \, \text{s} \):**
  \[
  v = v_0 + a \Delta t = 0 + 4 \times 2 = 8 \, \text{m/s}
  \]
  So, at \( t = 2 \, \text{s} \), the velocity is \( 8 \, \text{m/s} \).

- **From \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \):**
  Since the acceleration is zero in this interval, the velocity remains constant:
  \[
  v = 8 \, \text{m/s}
  \]
  So, at \( t = 4 \, \text{s} \), the velocity is still \( 8 \, \text{m/s} \).

- **From \( t = 4 \, \text{s} \) to \( t = 6 \, \text{s} \):**
  The acceleration is \( -2 \, \text{m/s}^2 \), so the velocity change is:
  \[
  v = 8 + (-2 \times 2) = 8 - 4 = 4 \, \text{m/s}
  \]
  Thus, at \( t = 6 \, \text{s} \), the velocity is \( 4 \, \text{m/s} \).

### 5. Summary of Results
- **Acceleration at \( t = 6 \, \text{s} \):** \( a = -2 \, \text{m/s}^2 \)
- **Velocity at \( t = 6 \, \text{s} \):** \( v = 4 \, \text{m/s} \)

Would you like more details or have any questions on the solution? 

Here are 5 related questions for further exploration:
1. What would the position of the object be at \( t = 6 \, \text{s} \)?
2. How would the velocity change if the mass were different?
3. What is the net work done on the object by the force over the time period?
4. Can we calculate the impulse experienced by the object during the first 2 seconds?
5. How does changing the direction of force after \( t = 4 \, \text{s} \) affect the motion?

**Tip:** When calculating velocity, it's important to account for each time interval separately if the force acting on the object changes.

Learn how to calculate the acceleration and velocity of a 1.0 kg object subjected to a step function force over time. Using Newton's Second Law and kinematic equations, we solve for the object's velocity at t = 6 s. Suitable for high school physics students.

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