Motion in a Straight Line

When a particle is released from rest position under gravity, then $v=0$ but $a \neq 0$

Equation of given line when velocity is plotted against position with intercept $=v_0$ and slope $=v_0 / x_0$ is,

$v=-\frac{v_0 x}{x_0}+v_0$ ...... (i)

($-$Ve sign is became the slope is decreasing) On differentiating with respect to $t$

$\frac{d v}{d t}=-\frac{v_0}{x_0} \frac{d x}{d t}+0$

or $\quad a=-\frac{v_0}{x_0} v$ ...... (ii)(

Putting $v$ from Eq. (i) in Eq. (ii),

$\begin{aligned} a & =-\frac{v_0}{x_0}\left[-\frac{v_0}{x_0} x+v_0\right] \\ & =\frac{v_0^2}{x_0^2} x-\frac{v_0^2}{x_0} \end{aligned}$

Which represents a straight line with positive slope and negative intercept.

We have

$S=u t+\frac{1}{2} a t^2$

$\begin{aligned} \text{or} \quad h & =(0) t+\frac{1}{2} \times 9.8 \times(4)^2 \quad\left(\begin{array}{l} \because a=g \\ \because u=0 \end{array}\right) \\ & =78.4 \mathrm{~m} \end{aligned}$

$\begin{gathered} \text{Given,} \quad \Delta t=0.2 \mathrm{~s} \\ t=4 \mathrm{~s} \end{gathered}$

Now, for error

$\frac{\Delta h}{h}= \pm 2\left(\frac{\Delta t}{t}\right)= \pm 2\left(\frac{0.2}{4}\right)= \pm 0.1$

$\begin{aligned} \text{or} \quad \Delta h & = \pm 0.1 \times h \\ & = \pm 0.1 \times 78.4= \pm 7.84 \mathrm{~m} \end{aligned}$