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Distance vs Displacement

For any arbit...

Question

For any arbitrary motion in space, which of the following relations are true:
$A) {\to v}_{average} = (\frac{1}{2}) (\to v (t_{1}) + \to v (t_{2}))$
$B) {\to v}_{average} = \frac{[\to r (t_{2}) - \to r (t_{1})]}{(t_{2} - t_{1})}$
$C) \to v (t) = \to v (0) + \to a t$
D) $\to r (t) = \to r (0) + \to v (0) t + (\frac{1}{2}) \to a t^{2}$
$E) {\to a}_{average} = ⎡ ⎣ \frac{\to v (t_{2}) - \to v (t_{1})}{(t_{2} - t_{1})} ⎤ ⎦$

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Solution

A) Step 1: Recall “Average velocity in uniform acceleration".
The formula ${\to v}_{average} = ⎡ ⎣ \frac{\to v (t_{1}) + \to v (t_{2})}{2} ⎤ ⎦$ holds good only for uniformly accelerated motion.
We are given that the motion of the particle is arbitrary.
So, the relation is not true.

B) Step 1 : Recall how to find average velocity.
Average velocity $({\to v}_{a v g}) = \frac{displacement}{time-interval}$
Average velocity $[{\to v}_{a v g}] = \frac{- \to Δ r}{Δ t}$
From figure : -
$- - \to A B = - - \to O B - - - \to O A$
$- \to Δ r = \to r (t_{2}) - \to r (t_{1})$
$∴ {\to v}_{a v g} = \frac{\to r (t_{2}) - \to r (t_{1})}{t_{2} - t_{1}}$
Hence, the relation is true.

C) Step 1: Recall Newton’s first equation of motion and its limitations.
The relation is not true.
For a particle which has an initial velocity at $t = 0$ is $\to v (0)$ .
If particle is moving with uniform acceleration ‘a’ then the velocity of the particle at time t is :-
$\to v (t) = \to v (0) + \to a t$
We have been given that motion is arbitrary in space,
so this relation is not valid here.

D) Step 1: Recall Newton’s second equation of motion and its limitation.
The relation is not true.
From Newton’s second equation of motion :-
$\to s = \to u t + \frac{1}{2} \to a t^{2}$
$\to r (t) - \to r (0) = \to v (0) t + \frac{1}{2} \to a t^{2}$
Where s is the displacement in time ‘t’,
$\to v (0) = \to u$ = initial velocity.
The motion of particle is arbitrary; This formula holds good only for uniformly accelerated motion.

E) Step 1 : Recall ${\to a}_{average} = \frac{Δ \to v}{Δ t}$
The relation is true.
For any arbitrary or uniform motion,
average acceleration
$[{\to a}_{a v g}] = \frac{Δ \to v}{Δ t} = \frac{\to v (t_{2}) - \to v (t_{1})}{t_{2} - t_{1}}$

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(The ‘average’ stands for average of the quantity over the time interval t₁ to t₂)

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