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 options is/are:

1.	(a) and (d) only	2.	(a) and (c) only
3.	(b) and (c) only	4.	(a) and (b) only

Hint:

v = \frac{Δ r}{Δ t}

$v= \frac{ \Delta r} { \Delta t}$ and

a_{a v g} = \frac{Δ v}{Δ t}

$a_{avg} = \frac{ \Delta v}{\Delta t}$

Step 1: Find the average velocity of the object.
If an object undergoes a displacement

Δ r

$\Delta r$ in time

Δ t,

$\Delta t,$ its average velocity is given by;

\Rightarrow v_{a v g} = \frac{Δ r}{Δ t} = \frac{r (t_{2}) - r (t_{1})}{t_{2} - t_{1}}

$\Rightarrow v_{avg} = \frac{\Delta r}{\Delta t}=\frac{r(t_2)-r(t_1)}{t_2-t_1}$
where

r_{1}

$r_1$ and

r_{2}

$r_2$ are position vectors corresponding to time

t_{1}

$t_1$ and

t_{2} .

$t_2.$

Step 2: Find the average acceleration.
The velocity of an object changes from

v_{1},

$v_1,$ to

v_{2}

$v_2$ in time

Δ t .

$\Delta t.$ Average acceleration is given by;

\Rightarrow a_{a v g} = \frac{Δ v}{Δ t} = \frac{v (t_{2}) - v (t_{1})}{t_{2} - t_{1}}

$\Rightarrow a_{avg} =\frac{\Delta v}{\Delta t}=\frac{v(t_2)-v(t_1)}{t_2- t_1}$
Hence, option (2) is the correct answer.