Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one $(s)$ :

{\to v}_{a v} = \frac{\to r (t_{2}) - \to r (t_{1})}{t_{2} = t_{1}}

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{\to v}_{a v} = \frac{1}{2} [\to v (t_{1}) + \to v (t_{2})]

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\to r = \frac{1}{2} (\to v (t_{2}) - \to v (t_{1})) (t_{2} - t_{1})

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{\to a}_{a v} = \frac{\to v (t_{2}) - \to v (t_{1})}{t_{2} - t_{1}}

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Solution

The correct option is C $\to r = \frac{1}{2} (\to v (t_{2}) - \to v (t_{1})) (t_{2} - t_{1})$
Step 1: Estimate the relation for average velocity.

Formula used: ${\to v}_{a v} = \frac{Δ \to r}{Δ t}$
Let the displacement of the particle is ${\to r}_{1}$ at time $t_{1}$ and $r_{2}$ at time $t_{2}$ .
So, average velocity
${\to v}_{a v g} = \frac{Δ \to r}{Δ t} = \frac{{\to r}_{2} - {\to r}_{1}}{t_{2} - t_{1}}$
When acceleration is uniform,
${\to v}_{a v} = \frac{1}{2} [\to v (t_{1}) + \to v (t_{2})]$
When acceleration is non-uniform,
${\to v}_{a v} \neq \frac{1}{2} [\to v (t_{1}) + \to v (t_{2})]$

Step 2: Estimate the relation for average acceleration.
Formula used: ${\to a}_{a v} = \frac{Δ \to v}{Δ \to t}$
Let the velocity of an object changes from ${\to v}_{1}$ to ${\to v}_{2}$ in time $Δ t$ , average acceleration
${\to a}_{a v} = \frac{Δ \to v}{Δ \to t} = \frac{\to v (t_{2}) - \to v (t_{1})}{t_{2} - t_{1}}$

Step 3: Estimate the expression for displacement.
Formula used: $\to r = \to u t + \frac{1}{2} \to a t^{2}$
As we know displacement is given by, $\to r = \to u t + \frac{1}{2} \to a t^{2}$
Here $\to u =$ intial velocity of the particle
$\to r = \to u t + \frac{1}{2} \frac{\to v (t_{2}) - \to v (t_{1})}{t_{2} - t_{1}} {(t_{2} - t_{1})}_{2}^{2}$
$\to r = \to u t + \frac{1}{2} (\to v (t_{2}) - \to v (t_{1})) (t_{2} - t_{1})$

Option $(c)$ will be correct only if initial velocity of the particle is zero and a is constant.

Final answer: $(a), (c)$

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