Question

In Figure, time and distance graph of a linear motion is given. Two positions of time and distance are recorded as, when $T=0,D=2$ and when $T=3,D=8.$ Using the concept of slope, find law of motion, i.e., how distance depends upon time.

Answer 1

Let $A=(0,2),B=(3,8),C=(T,D)$
We know that slope of a line through the points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$
Points $A,B\&C$ lie on the line.
So, $A,B\&C$ are collinear.
$\therefore \text{Slope of}AB=\text{Slope of}BC$
$\Rightarrow \frac{8-2}{3-0}=\frac{D-8}{T-3}$
$\Rightarrow \frac{6}{3}=\frac{D-8}{T-3}$
$\Rightarrow 2=\frac{D-8}{T-3}$
$\Rightarrow 2(T-3)=D-8$
$\Rightarrow 2T-6=D-8$
$\Rightarrow D-8=2T-6$
$\Rightarrow D=2T-6+8$
$\Rightarrow D=2T+2$
$\Rightarrow D=2(T+1)$

Hence, Distance depends on time as $D=2(T+1)$