Position Time Graph in Uniform Motion

To study the position-time graph of the uniform motion, let us again consider the example of a train moving along a straight track with a constant velocity of 15m/s. Suppose that at t = 0, the displacement of the train is 10m from the origin of the position axis. If $ \displaystyle {{{x}_{\circ }}}$ represents the displacement of the train at t = 0, then mathematically, we may say that

when $ \displaystyle t=0,{{x}_{\circ }}=10\text{m}$

The displacement of the train from the origin of the position axis at time t = 1s will be 10 + 15(1) i.e. 25m. Similarly, the displacement or position of the train at $ \displaystyle t=2s,3s,4s\text{ and }5s$ can be calculated.

If we plot a graph between time and position of the train from the data, the position-time graph will be like:

It follows that the position-time graph for an object in uniform motion along a straight path is a straight line inclined to the time-axis.

Let us now find the slope of the position time graph of an object in uniform motion. Since the graph is a straight line, its slope may be found between any two points on the graph. In the figure A and B are two points on position time graph corresponding to time t and t’. The position of the object (i.e. displacement from the origin of position axis) at time t is x (= AA’) and at time t’ is x’ (= BB’), so that $ \displaystyle AC={t}’-t\text{ and }BC={x}’-x$. Now, from right angled triangle ACB,

$ \displaystyle \text{slope of the }x-t\text{ graph }=\frac{{BC}}{{AC}}=\frac{{{x}’-x}}{{{t}’-t}}$

Hence, from the above equations, it follows that the velocity of uniform motion is equal to the slope of position time graph.