Question

A body is moving along a straight line with initial velocity v₀. Its acceleration a is constant. After t seconds, its velocity becomes v. The average velociy of the body over the given time interval is

• A. \(\bar{v}=\frac{v^2+{v_0}^2}{2at}\)
• B. \(\bar{v}=\frac{v^2+{v_0}^2}{at}\)
• C. \(\bar{v}=\frac{v^2-{v_0}^2}{2at}\)
• D. \(\bar{v}=\frac{v^2-{v_0}^2}{at}\)

Answer 1

The Correct Option is C

Given: 

• Initial velocity: \( v_0 \)
• Final velocity after \( t \) seconds: \( v \)
• Acceleration: \( a \) (constant)

Step 1: Using Kinematic Equation

We use the third equation of motion:

\[ v^2 = v_0^2 + 2 a s \]

Solving for \( s \) (displacement):

\[ s = \frac{v^2 - v_0^2}{2a} \]

Step 2: Average Velocity Formula

Average velocity is given by:

\[ \bar{v} = \frac{\text{total displacement}}{\text{total time}} \]

\[ \bar{v} = \frac{s}{t} = \frac{v^2 - v_0^2}{2at} \]

Step 3: Conclusion

The correct answer is:

\[ \bar{v} = \frac{v^2 - v_0^2}{2at} \]

Answer: The correct option is C.

Answer 2

To find the average velocity, we use the formula for average velocity when acceleration is constant: \[ \bar{v} = \frac{v + v_0}{2} \] where:
\( v_0 \) is the initial velocity,
\( v \) is the final velocity after time \( t \).

Now, using the kinematic equation: \[ v^2 = v_0^2 + 2a \cdot d \] where:
\( v_0 \) is the initial velocity,
\( v \) is the final velocity,
\( a \) is the acceleration,
\( d \) is the distance traveled.

Thus, the average velocity formula is derived from the kinematic equation. We calculate the average velocity over the time interval \( t \) as: \[ \bar{v} = \frac{v^2 - v_0^2}{2at} \]

Therefore, the correct answer is (C).