Answer :
a = ∆v/∆t
Say velocity is on the y-axis and time in on the x-axis. If you take the points (2,1) and (5,8) for example, the difference between the velocity points is 8-1 = 7. This difference is the change in velocity (i.e. ∆v, ∆ means change in) between these two points. The same goes for time: 5-2 = 3.
So we've concluded that in this example, ∆v = 7 and ∆t = 3. Now put these into the equation above to find acceleration:
7/3 = 2.333...
So the acceleration between these two points on the graph is 2.333...(recurring), or as a fraction, 7/3.
Say velocity is on the y-axis and time in on the x-axis. If you take the points (2,1) and (5,8) for example, the difference between the velocity points is 8-1 = 7. This difference is the change in velocity (i.e. ∆v, ∆ means change in) between these two points. The same goes for time: 5-2 = 3.
So we've concluded that in this example, ∆v = 7 and ∆t = 3. Now put these into the equation above to find acceleration:
7/3 = 2.333...
So the acceleration between these two points on the graph is 2.333...(recurring), or as a fraction, 7/3.
The magnitude of acceleration is
(change in speed) divided by (time for the change) .
Between the two points on the graph ...
-- Find the change in speed. (usually the y-axis)
-- Find the change in time. (usually the x-axis)
Divide (change in time) by (change in speed) between the two points.
That quotient is the average magnitude of acceleration during that time.
If the graph happens to be a straight line, then the magnitude of acceleration
is just the slope of the line.
(change in speed) divided by (time for the change) .
Between the two points on the graph ...
-- Find the change in speed. (usually the y-axis)
-- Find the change in time. (usually the x-axis)
Divide (change in time) by (change in speed) between the two points.
That quotient is the average magnitude of acceleration during that time.
If the graph happens to be a straight line, then the magnitude of acceleration
is just the slope of the line.
