The question involves a particle starting from rest and moving with an acceleration graph given in the image. Let's analyze the problem step by step.

Part (a) Determine the particle's speed at $t = 10.0 \, s$ and $t = 20.0 \, s$.

From the acceleration-time graph, we can calculate the velocity at specific times by finding the area under the acceleration-time curve. The velocity is the integral of acceleration over time:

$v(t) = v(0) + \int_0^t a(t') dt'$

Since the particle starts from rest ($v(0) = 0$), the velocity at any time will just be the area under the acceleration curve up to that time.

Velocity at $t = 10.0 \, s$

• The acceleration between $0 \, s$ and $10 \, s$ is constant at $2 \, \text{m/s}^2$.
• The area under the graph from $0 \, s$ to $10 \, s$ is a rectangle with height $2 \, \text{m/s}^2$ and base $10 \, s$, which gives:

$\text{Area} = 2 \times 10 = 20 \, \text{m/s}$

Thus, the velocity at $t = 10.0 \, s$ is $v(10.0) = 20 \, \text{m/s}$.

Velocity at $t = 20.0 \, s$

• From $t = 10 \, s$ to $t = 15 \, s$, the acceleration is $0$, so the velocity remains constant at $20 \, \text{m/s}$.
• From $t = 15 \, s$ to $t = 20 \, s$, the acceleration is $-3 \, \text{m/s}^2$. The area of this section is a rectangle with height $-3 \, \text{m/s}^2$ and base $5 \, s$, giving:

$\text{Area} = -3 \times 5 = -15 \, \text{m/s}$

Thus, the velocity at $t = 20.0 \, s$ is:

$v(20.0) = 20 \, \text{m/s} + (-15 \, \text{m/s}) = 5 \, \text{m/s}$

Part (b) Determine the distance traveled in the first 20 seconds.

To find the distance, we need to calculate the area under the velocity-time graph (which we can construct from the acceleration graph).

1. From $t = 0 \, s$ to $t = 10 \, s$, the velocity increases linearly from $0 \, \text{m/s}$ to $20 \, \text{m/s}$. This forms a triangle, and its area is:

$\text{Area} = \frac{1}{2} \times 10 \times 20 = 100 \, \text{m}$

2. From $t = 10 \, s$ to $t = 15 \, s$, the velocity remains constant at $20 \, \text{m/s}$, so the area is:

$\text{Area} = 20 \times 5 = 100 \, \text{m}$

3. From $t = 15 \, s$ to $t = 20 \, s$, the velocity decreases linearly from $20 \, \text{m/s}$ to $5 \, \text{m/s}$, forming a trapezoid. The area is:

$\text{Area} = \frac{1}{2} \times (20 + 5) \times 5 = \frac{1}{2} \times 25 \times 5 = 62.5 \, \text{m}$

Thus, the total distance traveled in the first $20 \, s$ is:

$\text{Total distance} = 100 + 100 + 62.5 = 262.5 \, \text{m}$

So the correct answer for the distance is $262.5 \, \text{m}$, not $300 \, \text{m}$ as shown in the image.

Would you like a more detailed breakdown or explanation of these calculations?

Here are 5 related questions to explore further:

1. How do you calculate velocity from a variable acceleration function?
2. What are the implications of negative acceleration on motion?
3. How does the area under an acceleration-time graph relate to the change in velocity?
4. How would the problem change if the initial velocity wasn't zero?
5. What is the difference between speed and velocity in problems like this?

Tip: In kinematic problems, always check whether acceleration is constant, as this greatly simplifies the process of finding velocity and displacement.