### 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. A line parallel to the $$x$$-axis in the velocity-time graph shows a car moving with a uniform velocity of $$40\ km\ h^{–1}$$. 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 the perpendicular lines from the points corresponding to 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 the 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.  