Theory:

A velocity-time graph shows the change in the velocity of an object in a straight line with respect to time.
Time is the quantity taken along the $$x$$-axis, and velocity is taken along the $$y$$-axis. The height of the velocity-time graph does not change with time for an object moving at a uniform velocity. The line parallel to the $$x$$-axis in the velocity-time graph shows a car moving with a uniform velocity of $$40\ km\ h^{–1}$$.

Velocity-time graph

The displacement of an object moving with uniform velocity is the product of velocity and time. The magnitude of displacement is equal to the area enclosed by the velocity-time graph and the time axis in the graph.

The distance travelled by car between time $$t_{1}$$ and $$t_{2}$$ can be found by drawing perpendicular lines from the points corresponding to the time $$t_{1}$$ and $$t_{2}$$ on the graph.

The height $$AC$$ or $$BD$$ represents the velocity of $$40\ km\ h^{–1}$$, and the length $$AB$$ represents the time ($$t_{2}\ – t_{1}$$).

Distance ($$s$$) moved by the car in time ($$t_{2}\ – t_{1}$$) is given as,

Plotting velocity–time graph:
Imagine a car being tested for its engine on a straight line. Assume that a person next to the driver records the car's velocity every $$4\ seconds$$ by noting the speedometer reading. The car's velocity in $$kilometres\ per\ hour$$ ($$km\ h^{–1}$$) and $$milliseconds\ per\ second$$ ($$m\ s^{–1}$$) at various points in time is given in the table.

 Time ($$s$$) Velocity ($$m s^-1$$) Velocity ($$km h^-1$$) $$0$$ $$0$$ $$0$$ $$4$$ $$2.5$$ $$9$$ $$8$$ $$5.0$$ $$18$$ $$16$$ $$7.5$$ $$27$$ $$20$$ $$10.0$$ $$36$$ $$24$$ $$12.5$$ $$45$$ $$28$$ $$15.0$$ $$54$$

Velocity-time graph for the motion of a car

It is clear from the graph that velocity changes by equal amounts in equal time intervals. Hence, the velocity-time graph for all uniformly accelerated motion is a straight line. The distance covered by the car can also be determined from the velocity-time graph. The shaded area under the velocity-time graph denotes the magnitude of displacement (distance) travelled by car in a given time interval.

The shaded area $$ABCD$$ under the graph shows the distance travelled by car with uniform velocity. The area $$ABCDE$$ under the velocity-time graph shows the distance ($$s$$) travelled by car.

Since the magnitude of the car's velocity changes as it accelerates, the distance ($$s$$) travelled will be determined by the region $$ABCDE$$ under the velocity-time graph.

$\begin{array}{l}s\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{area}\phantom{\rule{0.147em}{0ex}}\mathit{ABCDE}\\ \\ =\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{area}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{rectangle}\phantom{\rule{0.147em}{0ex}}\mathit{ABCD}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\mathit{area}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{triangle}\phantom{\rule{0.147em}{0ex}}\mathit{ADE}\\ \\ =\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{AB}\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\mathit{BC}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\frac{1}{2} \left(\mathit{AD}\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\mathit{DE}\right)\end{array}$

But, it differs in the case of non-uniform acceleration. The velocity-time graphs can have any shape in this type of motion.
Straight line velocity-time graph of an object

Irregularly shaped velocity-time graph of an object