The area under the graph (the region shaded in gray) is given by height ´
width, which is equivalent to velocity ´ time, which as you can see gives
the result of 8 meters. This is the same result from using the equation
Dx = vDt.
Figure 1.30 shows the example from earlier where a vehicle accelerates
from rest with a constant acceleration of 2 m/s
2
. Let’s say we’d like to cal

culate the distance traveled between the times t = 1 and t =3.
We know that the distance traveled between t = 1 and t = 3 is the area
beneath the graph between those times. As is clearly shown in the figure,
this is the sum of the areas of rectangle A and triangle B.
The area of A is given by the time displacement, t, multiplied by the
starting velocity, u, written as:
(1.83)
The area of B, a triangle, is half the area of the rectangle described by the
sides of the triangle. The sides of the triangle are given by the time dis

placement, t, and the difference between the finish velocity and the start
velocity, v – u. This can be written as:
(1.84)
Therefore, the total area under the graph between times t = 1 and t =3,
which is equivalent to the distance traveled, is the sum of these two terms,
given as:
(1.85)
A Math and Physics Primer  35
Physics
Figure 1.30
()Area A t u=D ´
1
() ( )
2
Area B v u t=D
1
()
2
xut vutD=D+  D