The problem you've uploaded involves the analysis of a velocity-time graph, $v(t)$, to determine several key quantities:

1. Total distance traveled (a),
2. Average velocity (b),
3. Acceleration-time graph (c).

Let's break down the steps:

Part (a): Calculate the total distance traveled

The distance is the area under the velocity-time graph. The graph is made up of several geometric shapes (trapezoids and rectangles). To find the total distance:

1. From $t = 0$ to $t = 3$ seconds, the graph forms a rectangle:

   • Velocity = 2 m/s, Time = 3 s.
   • Distance = $2 \, \text{m/s} \times 3 \, \text{s} = 6 \, \text{m}$.

2. From $t = 3$ to $t = 5$ seconds, it forms a trapezoid:

   • Base 1 (left side) = 2 m/s, Base 2 (right side) = 4 m/s, Time = 2 s.
   • Area of trapezoid = $\frac{1}{2} \times (2 + 4) \times 2 = 6 \, \text{m}$.

3. From $t = 5$ to $t = 7$ seconds, it forms a rectangle:

   • Velocity = 4 m/s, Time = 2 s.
   • Distance = $4 \, \text{m/s} \times 2 \, \text{s} = 8 \, \text{m}$.

4. From $t = 7$ to $t = 10$ seconds, it forms another trapezoid:

   • Base 1 (left side) = 4 m/s, Base 2 (right side) = 0 m/s, Time = 3 s.
   • Area of trapezoid = $\frac{1}{2} \times (4 + 0) \times 3 = 6 \, \text{m}$.

Total distance = $6 + 6 + 8 + 6 = 26 \, \text{m}$.

Part (b): Calculate the average velocity

The average velocity $\bar{v}$ can be calculated as:

$\bar{v} = \frac{\text{total distance}}{\text{total time}} = \frac{26 \, \text{m}}{10 \, \text{s}} = 2.6 \, \text{m/s}.$

Part (c): Draw the acceleration-time graph

To find the acceleration, we need to calculate the slope of the velocity-time graph in each segment:

1. From $t = 0$ to $t = 3$ seconds: The velocity is constant at 2 m/s, so acceleration = 0 m/s².
2. From $t = 3$ to $t = 5$ seconds: The velocity increases from 2 m/s to 4 m/s over 2 seconds.
   • Acceleration = $\frac{4 - 2}{5 - 3} = 1 \, \text{m/s}^2$.
3. From $t = 5$ to $t = 7$ seconds: The velocity is constant at 4 m/s, so acceleration = 0 m/s².
4. From $t = 7$ to $t = 10$ seconds: The velocity decreases from 4 m/s to 0 m/s over 3 seconds.
   • Acceleration = $\frac{0 - 4}{10 - 7} = -\frac{4}{3} \, \text{m/s}^2$.

The acceleration-time graph will have:

• A horizontal line at 0 m/s² from $t = 0$ to $t = 3$,
• A line at 1 m/s² from $t = 3$ to $t = 5$,
• A horizontal line at 0 m/s² from $t = 5$ to $t = 7$,
• A line at $-\frac{4}{3}$ m/s² from $t = 7$ to $t = 10$.

Would you like further clarification or any specific details on these steps?

5 Related Questions:

1. How is the displacement different from the total distance traveled in this scenario?
2. What would the graph look like if the object returned to its starting position?
3. How can we use calculus to verify the areas under the curve more accurately?
4. If the graph had been curved rather than linear, how would that affect the acceleration graph?
5. How does negative acceleration (deceleration) affect the motion of the object?

Tip: When analyzing motion graphs, remember that the slope of a velocity-time graph represents acceleration, while the area under the graph represents displacement or distance.