The figure gives the acceleration a versus time t for a particle moving along an x axis. The a-axis scale is set by a, 15.0 m/s². Att--20 s, the particle's velocity is 10.0 m/s. What is its velocity at t = 6.0 s?

To find the velocity of the particle at \( t = 6.0 \, \text{s} \), we can integrate the acceleration function over time to find the change in velocity. However, since no explicit function for the acceleration \( a(t) \) is provided, we'll need to analyze the figure (which is not currently visible to me). Based on your description, it sounds like we are working with an acceleration versus time graph.

Here is the general approach to solving the problem:

1. **Understand the given data**:
   - The velocity at \( t = -20 \, \text{s} \) is \( v(-20) = 10.0 \, \text{m/s} \).
   - The acceleration graph likely shows acceleration as a function of time between \( t = -20 \, \text{s} \) and \( t = 6.0 \, \text{s} \).
   - The acceleration axis scale is set by \( a = 15.0 \, \text{m/s}^2 \).

2. **Use the relationship between velocity and acceleration**:
   The velocity at any time \( t \) is given by:
   \[
   v(t) = v(t_0) + \int_{t_0}^{t} a(t') \, dt'
   \]
   where \( t_0 \) is the initial time, \( t \) is the final time, and \( a(t) \) is the acceleration.

3. **Interpret the acceleration graph**:
   Without seeing the graph, I can't provide exact steps, but the general approach would be:
   - Calculate the area under the \( a(t) \) curve between \( t = -20 \, \text{s} \) and \( t = 6.0 \, \text{s} \).
   - The area under the curve represents the change in velocity.

4. **Determine the velocity at \( t = 6.0 \, \text{s} \)**:
   - If the graph shows constant acceleration or linear changes in acceleration, we could compute the integral accordingly.

Please upload a picture of the graph if possible, and I'll be able to give you the exact solution based on the graphical information.

Would you like further details on any specific step? 

Here are 5 questions to help you expand on the concept:
1. How does the area under the acceleration vs. time graph relate to velocity?
2. What if the acceleration is a function of time, how does that affect velocity?
3. How do we account for negative acceleration in velocity calculations?
4. What are the units of the area under the acceleration-time graph?
5. Can we calculate velocity without knowing the exact function of acceleration?

**Tip:** The area under the acceleration-time graph between two points gives the change in velocity over that time interval.

The figure gives the acceleration a versus time t for a particle moving along an x axis. The a-axis scale is set by a = 15.0 m/s². At t = -20 s, the particle's velocity is 10.0 m/s. What is its velocity at t = 6.0 s?

Learn how to calculate the velocity of a particle at t = 6.0 s using the acceleration vs. time graph. Given initial conditions, solve the problem using kinematic equations and integration of acceleration over time. Suitable for students in Grades 11-12 and introductory physics courses.

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