Let's start by looking at a specific example of accelerated motion. Suppose an object moved forward along a strip of paper in the following manner: during the first second, it traveled a total of 0.1 meters; by the end of the second second it was located at 0.350 meters; then at 0.850 meters, 1.60 meters, 2.60 meters and finally by the end of six seconds, it was at 3.85 meters.
We will initially convert this information into a bar graph where each total distance is graphed against time. It is easily seen in the accompanying two position-time graphs that the object was covering greater and greater distances each second.
time
(seconds)
|
ending
position
(meters)
|
graphs |
| 1 |
0.100 |
If we look at the slopes of the tangent lines drawn to a curve for this position-time information we see that their slopes are all positive and get increasingly steeper. This indicates that the instantaneous velocity of the object was ever increasing. That is, the object was accelerating by gaining speed in a positive direction.
|
| 2 |
0.350 |
| 3 |
0.850 |
| 4 |
1.60 |
| 5 |
2.60 |
| 6 |
3.85 |
The first step in determining the magnitude of the object's average acceleration is to calculate how far it traveled during each time interval; that is, its displacement, and to create an appropriate velocity-time graph. In the following chart, we denote the midpoint and corresponding displacement during each time interval.
time
(seconds)
|
ending
position
(meters)
|
midpoint of
time interval
(seconds)
|
displacement
during time
interval
(meters)
|
Note: Red dots have been placed at the midpoint of each time interval.
|
| |
|
|
s = Δposition |
| 0 |
0 |
|
|
| |
|
0.5 |
0.100 |
| 1.0 |
0.100 |
|
|
| |
|
1.5 |
0.250 |
| 2.0 |
0.350 |
|
|
| |
|
2.5 |
0.500 |
| 3.0 |
0.850 |
|
|
| |
|
3.5 |
0.750 |
| 4.0 |
1.60 |
|
|
| |
|
4.5 |
1.00 |
| 5.0 |
2.60 |
|
|
| |
|
5.5 |
1.25 |
| 6.0 |
3.85 |
|
|
Since average velocity = displacement / time, we can easily complete one remaining table that will allow us to plot a velocity-time graph and determine the object's acceleration. Since the time intervals are always one second long, the average velocity during a time interval will equal the displacement during that same time interval.
time
(seconds)
|
ending
position
(meters)
|
midpoint of
time interval
(seconds)
|
displacement
during time
interval
(meters)
|
average velocity during time interval
(m/sec)
|
average acceleration
(m/sec2)
| | | | | s = Δposition | v = s/t
| a = Δv/Δt
| | 0 | 0 | | | | | | | | 0.5 | 0.100 | 0.100 | | | 1.0 | 0.100 | | | |
0.15 m/sec2 | | | | 1.5 | 0.250 | 0.250 | | | 2.0 | 0.350 | | | |
0.25 m/sec2 | | | | 2.5 | 0.500 | 0.500 | | | 3.0 | 0.850 | | | |
0.25 m/sec2 | | | | 3.5 | 0.750 | 0.750 | | | 4.0 | 1.60 | | | |
0.25 m/sec2 | | | | 4.5 | 1.00 | 1.00 | | | 5.0 | 2.60 | | | |
0.25 m/sec2 | | | | 5.5 | 1.25 | 1.25 | | | 6.0 | 3.85 | | | | |
Average acceleration equals the rate of change of velocity, or, a = Δv/Δt. In our example, the average acceleration equals a constant 0.25 m/sec2 with the exception of the first interval. This can be observed graphically by the fact that our line of best-fit on the following velocity vs time graph shown does not pass through the first data point.
We will now work backwards from our value for the average accleration to check our results. If we calculate the area under our acceleration vs time graph from t = 1.5 seconds until t = 5.5 seconds, we can determine the magnitude of the change in the object's velocity, Δv, during that time interval.
rectangle:
Δv = Areaa-t graph
Δv = [(5.5-1.5) sec] x [0.25 m/sec2]
Δv = 1 m/sec
| |
Examination of our graph of velocity vs time (red dots) shows us that the velocity changes from 0.25 m/sec at 1.5 seconds to 1.25 m/sec at 5.5 seconds, or 1 m/sec. Our results match!
Continuing to work backwards, if we examine the area under our velocity-time graph from 2 seconds until 5 seconds, we see that it is composed of a triangle sitting on top of a rectangle.
rectangular base:
s = Areav-t graph
s = [(5-2) sec] x [0.375 m/sec]
s = 1.125 meters
triangle:
s = Areav-t graph
s = ½[(5-2) sec] x [(1.125-0.375) m/sec]
s = 1.125 meters
| |
When we add these two areas together we get 1.125 meters + 1.125 meters or 2.25 meters. ALERT: It is unusual that the numerical values of these two areas are the same - do not make that assumption when calculating any future results.
Examination of our original position-time graph, shows the object's position at 2 seconds was 0.350 meters and at 6 seconds was 2.60 meters.
Its displacement between 2 and 5 seconds equaled 2.60 - 0.350 or 2.25 meters - exactly the same answer that we got when we calculated the areas on our velocity-time graph!
Summary: s-t graph |
s = Δposition
difference between
(yf - yo)
on the s-t graph |
slopes of tangent lines  |  area of v-t graph
| v-t graph |
Δv
difference between
(vf - vo)
on the v-t graph |
slope of v-t graph  |  area of a-t graph
| a-t graph | a-t graph |
|