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Finding Average Velocity

Statement - 1...

Question

Statement - 1 : For a particle undergoing rectilinear motion with constant acceleration, the average velocity cannot be equal for two different time intervals.
Statement -2 : The average velocity of particle moving with constant acceleration in a time interval is $\frac{\to u + \to v}{2}$ ,
where $\to u$ is the initial velocity and $\to v$ is the final velocity.

A

Both statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1.

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B

Both statement 1 and statement 2 are correct but statement 2 is not the correct explanation of statement 1.

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C

Statement 1 is correct and statement 2 is incorrect.

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D

Statement 1 is incorrect and statement 2 is correct.

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Solution

The correct option is D Statement 1 is incorrect and statement 2 is correct.
Average velocity of particle moving with constant acceleration in time interval $t = t_{1}$ to $t = t_{2}$ is equal to the average velocity in time interval $t = (t_{1} + Δ t)$ to $t = (t_{2} - Δ t)$ .
Let the initial velocity at time $t = 0$ be $u$
For interval $t = t_{1}$ to $t = t_{2}$
Velocity at time $t = t_{1}$ is given by
$v_{1} = u + a t_{1}$
Velocity at time $t = t_{2}$ is given by
$v_{2} = u + a t_{2}$
Average velocity in the time interval $t = t_{1}$ to $t = t_{2}$ is given by
$v_{a v g} = \frac{v_{1} + v_{2}}{2} = \frac{2 u + a (t_{1} + t_{2})}{2}$
For interval $t = t_{1} + Δ t$ to $t = t_{2} - Δ t$
Velocity at time $t = t_{1} + Δ t$ is given by
$v_{1}^{'} = u + a (t_{1} + Δ t)$
Velocity at time $t = t_{2} - Δ t$ is given by
$v_{2}^{'} = u + a (t_{2} - Δ t)$
Average velocity in the time interval $t = t_{1} + Δ t$ to $t = t_{2} - Δ t$ is given by
$v_{a v g} = \frac{v_{1}^{'} + v_{2}^{'}}{2} = \frac{u + a (t_{1} + Δ t) + u + a (t_{2} - Δ t)}{2} = \frac{2 u + a (t_{1} + t_{2})}{2}$
Hence , statement 1 is incorrect and statement 2 is correct

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