Let's begin by deriving the two kinematic equations shown in the image:

(1) Derivation of $v_f = v_i + at$

This is a standard kinematic equation, where:

• $v_f$ is the final velocity,
• $v_i$ is the initial velocity,
• $a$ is the acceleration, and
• $t$ is the time.

Step-by-step derivation:

1. Definition of acceleration: Acceleration is defined as the rate of change of velocity. $a = \frac{v_f - v_i}{t}$

2. Rearranging for $v_f$: Multiply both sides by $t$ to isolate $v_f$. $a t = v_f - v_i$

3. Solving for $v_f$: $v_f = v_i + at$

Thus, we derive: $v_f = v_i + at$

(2) Derivation of $\Delta x = \left( \frac{v_f + v_i}{2} \right) t$

This is another common kinematic equation, representing the displacement ($\Delta x$).

Step-by-step derivation:

1. Average velocity: The average velocity is defined as: $v_{\text{avg}} = \frac{v_f + v_i}{2}$

2. Relationship between displacement and average velocity: Displacement is the product of average velocity and time: $\Delta x = v_{\text{avg}} \cdot t$

3. Substitute the expression for average velocity: $\Delta x = \left( \frac{v_f + v_i}{2} \right) t$

Thus, we derive: $\Delta x = \left( \frac{v_f + v_i}{2} \right) t$

Would you like further details or have any questions? Here are five related questions that could deepen your understanding:

1. How can these equations be applied to motion with constant acceleration?
2. What would happen to these equations if the object was in free fall (where $a = g$)?
3. How do these equations change if the acceleration is zero?
4. Can you derive a third kinematic equation that eliminates time from the expressions?
5. How do we handle negative values for acceleration or velocity?

Tip: Always ensure units are consistent when solving kinematic problems (e.g., meters for distance, seconds for time).