Consider the following graph:
If we allow the ( y1 ) and ( x1 ) coordinates of the slope formula to be zero, a predictable slope begins to emerge as we move to the right. A steeper slope would necessitate a greater change in y-values relative to corresponding changes in x-values. Likewise, a less steep slope would indicate that changes in x-coordinates correspond to proportionally less changes in y-output values. If the y-coordinates are plots of an object’s position with x-coordinates reflecting a passage of time, the graph becomes one of an object’s velocity:
slope = ( rise / run ) = Δx/Δt = v
We may easily model changes in an object’s velocity over time with the same type of graph and corresponding equation:
slope = ( rise / run ) = Δv/Δt = a
Changes in an object’s velocity over a time interval give us the magnitude of its acceleration ( a ). Consider the graph of an accelerating object below:
After 2 seconds have transpired, the object is moving at 5 m/s. This means that it has traveled 10 meters at this time, but what distance was covered beforehand, and what distance is covered by the object afterwards? Since the product of velocity and time gave us the distance traveled at the 2 second interval, the total area under the graph represents the total distance covered by the object in question. The area under the curve from the 0-to-4 second interval is that of a triangle, and the area from the 4-to-6 second interval is the area of a rectangle:
Area triangle = ½ bh
Area rectangle = bh
Total distance traveled = ½ b1h1 + b2h2
Total distance = [ ( ½ )( 4s )( 10 m/s ) ] + ( 2s )( 10 m/s )
Total distance = 20 m + 20 m = 40 m
The object in question accelerated at 2.5 m/s2 over 4 seconds. After 4 seconds transpire, the object ceases accelerating and advances forward with a constant velocity of 10 m/s.
RENAISSANCE PHYSICS 