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.