[Q1] For an object undergoing uniformly accelerated linear motion, the key kinematic relationships are used to derive the proof.

1. Average Velocity ($\bar{v}$) The average velocity is defined as the total displacement ($\Delta x$) divided by the total time interval ($T$).

$$\bar{v} = \frac{\Delta x}{T}$$

For uniformly accelerated motion, the displacement can be expressed in terms of the initial and final velocities:

$$\Delta x = \frac{1}{2}(v_i + v_f)T$$

Substituting this into the definition of average velocity gives:

$$\bar{v} = \frac{\frac{1}{2}(v_i + v_f)T}{T} = \frac{1}{2}(v_i + v_f)$$

2. Instantaneous Velocity at Mid-Time ($v(T/2)$) The instantaneous velocity $v(t)$ at any time $t$ is given by:

$$v(t) = v_i + at$$

where $a$ is the constant acceleration. At the mid-time point, $t = T/2$, the velocity is:

$$v(T/2) = v_i + a\frac{T}{2}$$

The acceleration can be expressed in terms of the initial and final velocities over the interval $T$:

$$v_f = v_i + aT \implies a = \frac{v_f - v_i}{T}$$

Substituting this expression for acceleration into the equation for $v(T/2)$:

$$v(T/2) = v_i + \left(\frac{v_f - v_i}{T}\right)\frac{T}{2}$$ $$v(T/2) = v_i + \frac{v_f - v_i}{2} = \frac{2v_i + v_f - v_i}{2}$$ $$v(T/2) = \frac{1}{2}(v_i + v_f)$$

3. Conclusion By comparing the derived expressions for the average velocity and the instantaneous velocity at the mid-time point, we find they are identical.

$$\bar{v} = v(T/2) = \frac{1}{2}(v_i + v_f)$$

This completes the proof.

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Alternative Proof (using Calculus) The average velocity is the time-average of the instantaneous velocity function $v(t) = v_i + at$:

$$\bar{v} = \frac{1}{T} \int_{0}^{T} v(t) \,dt = \frac{1}{T} \int_{0}^{T} (v_i + at) \,dt$$ $$\bar{v} = \frac{1}{T} \left[ v_i t + \frac{1}{2}at^2 \right]_0^T = \frac{1}{T} \left( v_i T + \frac{1}{2}aT^2 \right) = v_i + \frac{1}{2}aT$$

The instantaneous velocity at the mid-time point $t=T/2$ is:

$$v(T/2) = v_i + a\left(\frac{T}{2}\right) = v_i + \frac{1}{2}aT$$

Thus, $\bar{v} = v(T/2)$.