Answer :
Answer:
The equation: Average velocity= ( initial velocity+ Final velocity) /2 will always be true when object moves with uniform acceleration
Answer:
Explanation:
To prove that the average velocity (\( v_{\text{avg}} \)) in uniform acceleration is equal to the average of the initial velocity (\( v_i \)) and the final velocity (\( v_f \)), we can use the basic definitions and equations of motion. Here’s a step-by-step proof:
1. **Definition of Average Velocity**:
The average velocity is defined as the total displacement divided by the total time:
\[
v_{\text{avg}} = \frac{\Delta x}{\Delta t}
\]
where \(\Delta x\) is the displacement and \(\Delta t\) is the time interval.
2. **Equation of Motion for Uniform Acceleration**:
For an object under uniform acceleration, the displacement can be expressed using the equation of motion:
\[
\Delta x = v_i \Delta t + \frac{1}{2} a (\Delta t)^2
\]
where \( v_i \) is the initial velocity, \( a \) is the constant acceleration, and \( \Delta t \) is the time interval.
3. **Final Velocity Equation**:
The final velocity (\( v_f \)) under uniform acceleration is given by:
\[
v_f = v_i + a \Delta t
\]
4. **Calculate the Total Displacement**:
From the equation of motion, we have:
\[
\Delta x = v_i \Delta t + \frac{1}{2} a (\Delta t)^2
\]
5. **Express Displacement in Terms of Initial and Final Velocity**:
We know that \( v_f = v_i + a \Delta t \). Solving for \( a \Delta t \):
\[
a \Delta t = v_f - v_i
\]
Substitute this back into the displacement equation:
\[
\Delta x = v_i \Delta t + \frac{1}{2} (v_f - v_i) \Delta t
\]
Factor out \(\Delta t\):
\[
\Delta x = \Delta t \left( v_i + \frac{1}{2} (v_f - v_i) \right)
\]
Simplify the expression inside the parentheses:
\[
\Delta x = \Delta t \left( \frac{2v_i + v_f - v_i}{2} \right) = \Delta t \left( \frac{v_i + v_f}{2} \right)
\]
6. **Find Average Velocity**:
Recall that the average velocity is:
\[
v_{\text{avg}} = \frac{\Delta x}{\Delta t}
\]
Substitute \(\Delta x\) from the previous step:
\[
v_{\text{avg}} = \frac{\Delta t \left( \frac{v_i + v_f}{2} \right)}{\Delta t}
\]
The \(\Delta t\) terms cancel out:
\[
v_{\text{avg}} = \frac{v_i + v_f}{2}
\]
Therefore, we have proved that the average velocity under uniform acceleration is equal to the average of the initial and final velocities:
\[
v_{\text{avg}} = \frac{v_i + v_f}{2}
\]