https://file.aiquickdraw.com/mathbot/file/17254634387489qjs5mv3.jpg

Let's go through each part of the problem step by step.

### Part 1: Acceleration at 1 second
Acceleration is the rate of change of velocity over time. Mathematically, it's calculated using the formula:

\[
a = \frac{\Delta v}{\Delta t}
\]

From the graph:
- At \( t = 0 \) seconds, the velocity (\( v \)) is 0 m/s.
- At \( t = 2 \) seconds, the velocity (\( v \)) is 8 m/s.

So, between \( t = 0 \) and \( t = 2 \), the change in velocity \( \Delta v = 8 - 0 = 8 \) m/s, and the change in time \( \Delta t = 2 - 0 = 2 \) seconds.

The acceleration at \( t = 1 \) second (since the acceleration is constant between 0 and 2 seconds) is:

\[
a = \frac{8 \, \text{m/s}}{2 \, \text{s}} = 4 \, \text{m/s}^2
\]

**Answer:** The acceleration at 1 second is **4 m/s²**.

### Part 2: Velocity at 2 seconds
The velocity at 2 seconds can be read directly from the graph.

- At \( t = 2 \) seconds, the velocity (\( v \)) is **8 m/s**.

**Answer:** The velocity at 2 seconds is **8 m/s**.

### Part 3: Position at 2 seconds
To find the position at 2 seconds, we integrate the velocity over time, starting from the initial position. 

Since the initial position is given as 10 m, and the velocity changes uniformly from 0 to 8 m/s over 2 seconds, the position at \( t = 2 \) seconds is:

\[
x(2) = x_0 + \text{Area under the velocity-time graph from 0 to 2 seconds}
\]

The area under the graph from 0 to 2 seconds forms a right triangle with a base of 2 seconds and a height of 8 m/s.

\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 8 = 8 \, \text{m}
\]

So, the position at 2 seconds is:

\[
x(2) = 10 \, \text{m} + 8 \, \text{m} = 18 \, \text{m}
\]

**Answer:** The position at 2 seconds is **18 m**.

---

Would you like more details on any part of this solution, or do you have any additional questions? 

Here are some related questions you might find useful:
1. What is the acceleration between 2 and 5 seconds?
2. What is the velocity at 5 seconds?
3. What is the total distance traveled by the object by 9 seconds?
4. What is the position at 4 seconds?
5. How would the graph look if the object had a constant acceleration throughout the motion?

**Tip:** When analyzing motion, always consider the area under the velocity-time graph to find displacement.

Consider the plot below describing motion along a straight line with an initial position of 10 m.
1. What is the acceleration at 1 second?
2. What is the velocity at 2 seconds?
3. What is the position at 2 seconds?

Learn how to calculate acceleration, velocity, and position using a velocity-time graph. This problem involves determining acceleration at a specific time, reading velocity values, and computing position using initial conditions. Suitable for students in Grades 9-12.

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