Question

For any arbitrary motion in space, which of the following relations is true?(The average stands for average of the quantity over the time interval ($t_1$) to ($t_2$)

• A. ${\overrightarrow{v}}_{average}$ = 1/2 [$\overrightarrow{v}\left(t_1\right)$ + $\overrightarrow{v}\left(t_2\right)$]
• B. ${\overrightarrow{v}}_{average}$ = [$\overrightarrow{r}\left(t_2\right)-\overrightarrow{r}\left(t_1\right)\left]/\right(t_2)-(t_1$)
• C. $\overrightarrow{v}$(t)= $\overrightarrow{v}$0 + $\overrightarrow{a}$t
• D. $\overrightarrow{r}$(t)= $\overrightarrow{r}$0 +$\overrightarrow{v}$0t +1/2 $\overrightarrow{a}$ $t^2$

Answer 1

The correct option is B ${\overrightarrow{v}}_{average}$ = [$\overrightarrow{r}\left(t_2\right)-\overrightarrow{r}\left(t_1\right)\left]/\right(t_2)-(t_1$)

$v_{avg}=\frac{Totaldisplacement}{Totaltime}$

If arbitrary motional function $x\left(t\right)$ is given.

Then, average speed in interval $t_1\le t\le t_2$

$=x\left(t_2\right)-\frac{x\left(t_1\right)}{(t_2-t_1)}$

Similarly average acceleration$=V\left(t_2\right)-\frac{V\left(t\right)}{(t_2-t_1)}$

The relation (b) is true, others are false because relations (a), (c) and (d) hold only for uniformly accelerated motion.