https://file.aiquickdraw.com/mathbot/file/1726799289010j4thzdwp.jpg

From the image, we are given the acceleration vs. time graph of a particle, and we are tasked with determining:

1. The particle's speed at \( t = 10.0 \, s \) and \( t = 20.0 \, s \).
2. The distance traveled by the particle in the first \( 20.0 \, s \).

### Part (a) – Determining Speed at \( t = 10.0 \, s \) and \( t = 20.0 \, s \)

The particle starts from rest, meaning the initial velocity at \( t = 0 \, s \) is \( v_0 = 0 \, m/s \). To find the velocity at different times, we can integrate the acceleration over time.

1. **From \( t = 0 \, s \) to \( t = 10.0 \, s \):**

   The acceleration is \( a = 1 \, m/s^2 \). The change in velocity \( \Delta v \) over this interval can be calculated as:

   \[
   \Delta v = a \times \Delta t = 1 \, m/s^2 \times 10 \, s = 10 \, m/s
   \]

   Since the initial velocity \( v_0 = 0 \, m/s \), the velocity at \( t = 10 \, s \) is:

   \[
   v(10 \, s) = 10 \, m/s
   \]

2. **From \( t = 10.0 \, s \) to \( t = 15.0 \, s \):**

   The acceleration is \( a = 0 \, m/s^2 \), meaning the velocity remains constant. Hence, at \( t = 15 \, s \), the velocity is still \( v(15 \, s) = 10 \, m/s \).

3. **From \( t = 15.0 \, s \) to \( t = 20.0 \, s \):**

   The acceleration is \( a = -2 \, m/s^2 \). The change in velocity can be calculated as:

   \[
   \Delta v = -2 \, m/s^2 \times 5 \, s = -10 \, m/s
   \]

   Therefore, the velocity at \( t = 20 \, s \) is:

   \[
   v(20 \, s) = 10 \, m/s - 10 \, m/s = 0 \, m/s
   \]

So, the velocity at \( t = 10.0 \, s \) is \( 10.0 \, m/s \), and at \( t = 20.0 \, s \) it is \( 0.0 \, m/s \).

### Part (b) – Distance Traveled in the First 20.0 Seconds

To determine the distance, we need to calculate the area under the acceleration-time graph, which gives us the velocity, and use this to find the distance. Alternatively, we can use the velocity-time graph.

1. **From \( t = 0 \, s \) to \( t = 10.0 \, s \):**

   The velocity increases linearly from \( 0 \, m/s \) to \( 10 \, m/s \). The distance traveled in this period is:

   \[
   d_1 = \frac{1}{2} \times v_{\text{max}} \times t = \frac{1}{2} \times 10 \, m/s \times 10 \, s = 50 \, m
   \]

2. **From \( t = 10.0 \, s \) to \( t = 15.0 \, s \):**

   The velocity is constant at \( 10 \, m/s \). The distance traveled is:

   \[
   d_2 = v \times t = 10 \, m/s \times 5 \, s = 50 \, m
   \]

3. **From \( t = 15.0 \, s \) to \( t = 20.0 \, s \):**

   The velocity decreases linearly from \( 10 \, m/s \) to \( 0 \, m/s \). The distance traveled is:

   \[
   d_3 = \frac{1}{2} \times v_{\text{max}} \times t = \frac{1}{2} \times 10 \, m/s \times 5 \, s = 25 \, m
   \]

Thus, the total distance traveled in the first 20.0 seconds is:

\[
d_{\text{total}} = d_1 + d_2 + d_3 = 50 \, m + 50 \, m + 25 \, m = 125 \, m
\]

### Final Answers:
- \( v(10.0 \, s) = 10.0 \, m/s \)
- \( v(20.0 \, s) = 0.0 \, m/s \)
- Total distance traveled in the first \( 20.0 \, s \) is \( 125 \, m \).

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Would you like further details or clarifications on these calculations?

Here are five related questions to expand on the concepts:

1. How would the velocity change if the initial acceleration were different?
2. What would happen if the negative acceleration lasted longer than 5 seconds?
3. How does integrating acceleration relate to the concept of velocity?
4. Can you describe how to derive displacement using the velocity-time graph?
5. What real-life scenarios can this motion graph represent?

**Tip:** When working with acceleration-time graphs, always check if the acceleration is constant or changing. This helps to identify how velocity evolves over time.

A particle starts from rest and accelerates as shown in the acceleration-time graph. Determine: (a) The particle's speed at t = 10.0 s and t = 20.0 s. (b) The distance traveled in the first 20.0 s.

In this physics problem, learn how to calculate the particle's velocity at specific times and determine the total distance traveled using an acceleration-time graph. This example involves key kinematic equations and concepts such as velocity, acceleration, and distance.

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