Average Acceleration Calculation:

The formula for average acceleration is:

$a_{\text{av,x}} = \frac{v_{2x} - v_{1x}}{t_2 - t_1} = \frac{\Delta v_x}{\Delta t}$

Where:

• $v_{1x}$ is the initial velocity
• $v_{2x}$ is the final velocity
• $t_1$ and $t_2$ are the initial and final times respectively

Given that $t_1 = 2 \, \text{s}$ and $t_2 = 4 \, \text{s}$, the time interval $\Delta t = t_2 - t_1 = 2 \, \text{s}$.

Case-by-Case Calculation:

(a) $v_{1x} = 0.8 \, \text{m/s}, v_{2x} = 1.2 \, \text{m/s}$ (speeding up)

$a_{\text{av,x}} = \frac{1.2 \, \text{m/s} - 0.8 \, \text{m/s}}{2 \, \text{s}} = \frac{0.4 \, \text{m/s}}{2 \, \text{s}} = 0.2 \, \text{m/s}^2$

(b) $v_{1x} = 1.6 \, \text{m/s}, v_{2x} = 1.2 \, \text{m/s}$ (slowing down)

$a_{\text{av,x}} = \frac{1.2 \, \text{m/s} - 1.6 \, \text{m/s}}{2 \, \text{s}} = \frac{-0.4 \, \text{m/s}}{2 \, \text{s}} = -0.2 \, \text{m/s}^2$

(c) $v_{1x} = -0.4 \, \text{m/s}, v_{2x} = -1.0 \, \text{m/s}$ (speeding up)

$a_{\text{av,x}} = \frac{-1.0 \, \text{m/s} - (-0.4 \, \text{m/s})}{2 \, \text{s}} = \frac{-0.6 \, \text{m/s}}{2 \, \text{s}} = -0.3 \, \text{m/s}^2$

(d) $v_{1x} = -1.6 \, \text{m/s}, v_{2x} = -0.8 \, \text{m/s}$ (slowing down)

$a_{\text{av,x}} = \frac{-0.8 \, \text{m/s} - (-1.6 \, \text{m/s})}{2 \, \text{s}} = \frac{0.8 \, \text{m/s}}{2 \, \text{s}} = 0.4 \, \text{m/s}^2$

Fifth Set of Data:

For this set, we are given:

• Average acceleration $a_{\text{av,x}} = -0.7 \, \text{m/s}^2$
• Initial velocity $v_{1x} = -0.4 \, \text{m/s}$
• Time interval $\Delta t = 2 \, \text{s}$

We can use the average acceleration formula to solve for the final velocity $v_{2x}$:

$a_{\text{av,x}} = \frac{v_{2x} - v_{1x}}{\Delta t}$

Substituting the values:

$-0.7 = \frac{v_{2x} - (-0.4)}{2}$

$-0.7 \times 2 = v_{2x} + 0.4$

$-1.4 = v_{2x} + 0.4$

$v_{2x} = -1.8 \, \text{m/s}$

Summary of Results:

1. Case (a): $a_{\text{av,x}} = 0.2 \, \text{m/s}^2$
2. Case (b): $a_{\text{av,x}} = -0.2 \, \text{m/s}^2$
3. Case (c): $a_{\text{av,x}} = -0.3 \, \text{m/s}^2$
4. Case (d): $a_{\text{av,x}} = 0.4 \, \text{m/s}^2$
5. Fifth set: Final velocity $v_{2x} = -1.8 \, \text{m/s}$

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

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Further Questions:

1. What is the significance of positive vs negative acceleration?
2. How does changing the time interval affect the average acceleration?
3. How can we interpret negative velocities in these calculations?
4. What happens if the final velocity is zero in any of the above cases?
5. Can acceleration remain constant while velocity changes?

Tip: Remember that acceleration can be negative (deceleration) when an object slows down or moves in the opposite direction of its initial velocity.