Under constant acceleration, the average velocity of a particle is half the sum of its initial and final velocities. Is this still true if the acceleration is not constant? Explain.
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It is not still true that the average velocity of the particle is equal to half the sum of the initial and final velocities when the acceleration of the particle is not constant

Explanation:

The motion of a particle under constant acceleration, 'a', is be given by the following kinematic equations;

v² = u² + 2·a·s

v = u + a·t

Where;

v = The final velocity of the particle

u = The initial velocity of the particle

a = The acceleration of the particle

s = The distance through which the particle travels

t = The time of motion of the particle

By simplifying the above equation, we have;

v² - u² = 2·a·s

(v² - u²)/(2·a) = s

(v - u) × (v + u)/(2 × a) = s

((v - u)/a) × ((v + u)/2) = s

From v = u + a·t, we have;

t = (v - u)/a

∴ ((v - u)/a) × ((v + u)/2) = t × ((v + u)/2) = s

∴ ((v + u)/2) = s/t

The average velocity = (Total distance traveled by the particle) ÷ (The time of travel of the particle)

∴ The average velocity = s/t = ((v + u)/2) = Half the sum of the initial and final velocity

However, it is not still true that the average velocity of the particle is equal to half the sum of the initial and final velocities when the acceleration of the particle, 'a', is not constant, as the velocity time graph is no longer a straight line graph and the distance traveled by the particle, 's', which is the area under the velocity time graph, 'A', (given by the sum of area of the triangle and the rectangle given by the area under straight line graph for constant velocity) cannot be given directly by the product of the time and the average velocity.

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