Velocity Time Graph of Uniform Motion

Let us considered a train moving with a constant velocity of say 15m/s along a straight track so as to study the velocity time graph of the uniform motion of an object in a straight line. As a train is in uniform motion, the magnitude of its velocity at t = 0, t = 1s, t = 2s,… will always be 15m/s and therefore graph between time and velocity of the train will be shown in figure.

Hence, velocity time graph for an object in uniform motion along a straight path is a straight line parallel to time-axis.

The velocity time graph of an object may be used to calculate it’s displacement in a given time interval geometrically. In the above graph A and B are the two points on the velocity-time graph corresponding to the instants t and t’. If the motion of the object is with a constant velocity $ \displaystyle v$, then AA’ = BB’ = $ \displaystyle v$. Also, A’B’ = t’ – t. Therefore,

area under $ \displaystyle v-t$ graph between t and t’ = $ \displaystyle \text{area }AB{B}’A=A{A}’\times {A}'{B}’=v\left( {{t}’-t} \right)$

But, since $ \displaystyle v\left( {{t}’-t} \right)=\overrightarrow{{{x}’}}-\overrightarrow{x}$

Hence , $ \displaystyle \overrightarrow{{{x}’}}-\overrightarrow{x}=\text{area }AB{B}’A$

i.e. displacement of an object in time interval $ \displaystyle \left( {{t}’-t} \right)$ is numerically equal to the area under velocity-time graph between the instants t and t’.

Note: It may be pointed out that this geometrical method of finding the displacement of an object holds good even in the case, when the body is moving with negative velocity. In such a case, the area below the velocity-time graph is taken negative and corresponding to this the displacement will also be negative.