Interpreting a graph of an object’s acceleration can be tricky, especially when the graph is linear:
An object moving with a constant velocity moves through equal distance segments as time transpires. To the contrary, an object that is undergoing a constant acceleration has a velocity that changes as time transpires. As a consequence, plots of the distance traveled each second by an accelerating object would not be linear.
For example, consider the time interval from 2 seconds to 4 seconds. The object’s velocity changes from 20 m/s to 40 m/s. This is a net change of 20 m/s. If we multiply 20 m/s by 2 seconds, we get 40 meters. This answer is wrong! The net displacement that occurs during this time interval must be evaluated via the area of the curve underneath:
Furthermore,
a = ( v / t ) = 10 m/s2
The object’s velocity changes by 10 meters per second every second. Furthermore, the kinematic equation for displacement is as follows:
x = vot +/- ½ at2
Since our starting velocity is 20 m/s, the distance displaced over the 2 second to 4 second interval can be determined via variable substitution:
x = ( 20 m/s )( 2 s ) + ½ ( 10 m/s2 )( 2 s )2
x = 40 m + 20 m = 60 m
Likewise, we may determine the area of the square and triangular areas in the graph above and subsequently add them together to determine the displacement that occurred between the 20 second to 40 second interval:
As = bh = ( 20 m/s )( 2 s ) = 40 m
At = ½ bh = ½ ( 20 m/s )( 2 s ) = 20 m
As + At = 60 m
RENAISSANCE PHYSICS 