Ncert Exercises & Solutions

The following are four different relations about displacement, velocity and acceleration for the motion of a particle in general.

(a)	$v_{a v}=1 / 2\left[v\left(t_1\right)+v\left(t_2\right)\right]$
(b)	$v_{{av}}={r}\left({t}_2\right)-{r}\left({t}_1\right) / {t}_2-{t}_1$
(c)	$r=1 / 2\left[v\left(t_2\right)-v\left(t_1\right)\right]\left({t}_2-{t}_1\right)$
(d)	${a}_{{av}}=v\left({t}_2\right)-v\left({t}_1\right) / {t}_2-{t}_1$

The incorrect alternative/s is/are:

1.	(a), (d)	2.	(a), (c)
3.	(b), (c)	4.	(a), (b)

_{(2) Hint: Recall the concept of average velocity and average acceleration.}

_Step_{1: Find the average velocity of the object.}

_{If an object undergoes a displacement Ar in time At, its average velocity is given by

$v = \frac{Δ r}{Δ t} = \frac{r_{2} - r_{1}}{t_{2} - t_{1}}$ where $r_{1}$ and $r_{2}$ are position vectors corresponding to time t, and t}

_{Step 2: Find the average acceleration.

t the velocity of an object changes from v, to v in time At. Average acceleration is given by

$a_{a v} = \frac{Δ v}{Δ t} = \frac{v_{2} - v_{1}}{t_{2} - t_{2}}$

But, when acceleration is non-uniform,}

_{$\begin{array}{l} v_{av} \neq \frac{v_{1} + v_{2}}{2} \\ We can write Δv = \frac{Δr}{Δt} \\ Hence, Δr = r_{2} - r_{1} = (v_{2} - v_{1}) (t_{2} - t_{1}) \end{array}$}

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(a)	\(v_{a v}=1 / 2\left[v\left(t_1\right)+v\left(t_2\right)\right]\)
(b)	\(v_{{av}}={r}\left({t}_2\right)-{r}\left({t}_1\right) / {t}_2-{t}_1\)
(c)	\(r=1 / 2\left[v\left(t_2\right)-v\left(t_1\right)\right]\left({t}_2-{t}_1\right)\)
(d)	\({a}_{{av}}=v\left({t}_2\right)-v\left({t}_1\right) / {t}_2-{t}_1\)