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use some of this formula for me to understand it better

Great! Let's analyze the formulas on the blackboard and relate them to the problem you provided earlier. The formulas on the board are fundamental equations of motion under constant acceleration.

### Key Formulas on the Blackboard:

1. **Velocity and Acceleration Relationships**:
   \[
   v = \frac{d}{t} \quad \text{and} \quad a = \frac{\Delta v}{\Delta t}
   \]
   - \( v \) is velocity (m/s or km/h).
   - \( d \) is displacement (meters or kilometers).
   - \( t \) is time (seconds or hours).
   - \( a \) is acceleration (m/s² or km/h²).

2. **Displacement Formula**:
   \[
   d_f = d_i + v_i t + \frac{1}{2} a t^2
   \]
   - \( d_f \) is the final displacement.
   - \( d_i \) is the initial displacement.
   - \( v_i \) is the initial velocity.
   - \( t \) is time.
   - \( a \) is acceleration.

3. **Average Velocity**:
   \[
   \bar{v} = \frac{v_i + v_f}{2}
   \]
   - \( \bar{v} \) is the average velocity.
   - \( v_i \) is the initial velocity.
   - \( v_f \) is the final velocity.

4. **Final Velocity**:
   \[
   v_f = v_i + at
   \]
   - \( v_f \) is the final velocity.
   - \( v_i \) is the initial velocity.
   - \( a \) is acceleration.
   - \( t \) is time.

5. **Another Displacement Formula**:
   \[
   d = v_i (t_f - t_i) + \frac{1}{2} a (t_f - t_i)^2
   \]
   - This formula gives the displacement between two time intervals \( t_i \) (initial time) and \( t_f \) (final time).

### Applying to Your Problem:

Given the problem of a car with an initial velocity \( v_i = 30 \, \text{km/h} \), constant acceleration \( a = 2 \, \text{km/h}^2 \), and time interval \( t_i = 20 \) seconds and \( t_f = 60 \) seconds, we can use the **last formula** shown on the blackboard:

\[
d = v_i (t_f - t_i) + \frac{1}{2} a (t_f - t_i)^2
\]

### Step-by-Step Application:

1. **Convert time to hours**:
   \[
   t_i = \frac{20}{3600} \text{ hours}, \quad t_f = \frac{60}{3600} \text{ hours}
   \]

2. **Substitute values**:
   \[
   d = 30 \times (t_f - t_i) + \frac{1}{2} \times 2 \times (t_f - t_i)^2
   \]
   - Substitute \( t_f - t_i = \frac{40}{3600} \text{ hours} = \frac{1}{90} \text{ hours} \approx 0.0111 \text{ hours} \).

3. **Compute**:
   \[
   d \approx 30 \times 0.0111 + \frac{1}{2} \times 2 \times (0.0111)^2
   \]
   \[
   d \approx 0.333 + 0.00012 = 0.33312 \, \text{km}
   \]

So, the distance traveled by the car during this time frame is approximately **0.33312 km**.

Would you like any further clarification or details?

---

Here are 5 related questions:
1. How can we use these equations to find the final velocity at \( t = 60 \) seconds?
2. How does the initial displacement \( d_i \) affect the total distance?
3. What happens if the acceleration \( a \) changes during the time interval?
4. How would the calculations change if the car decelerated instead of accelerating?
5. Can these equations be used for non-linear acceleration? Why or why not?

**Tip:** Always check if your time units are consistent with your velocity and acceleration units before solving any kinematic problem!

Explore how to apply kinematic equations of motion with constant acceleration to solve problems. Learn about velocity and displacement formulas, including step-by-step examples with specific values. Suitable for advanced high school students and physics enthusiasts.

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