 Accelerated Motion: A Data Analysis Approach Printer Friendly Version
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) endingposition(meters) midpoint oftime interval(seconds) displacementduring timeinterval(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) endingposition(meters) midpoint oftime interval(seconds) displacementduring timeinterval(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 graphΔv = [(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 graphs = [(5-2) sec] x [0.375 m/sec] s = 1.125 meters triangle: s = Areav-t graphs = 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 = Δpositiondifference between(yf - yo)on the s-t graph slopes of tangent lines  area of v-t graph v-t graph Δvdifference between(vf - vo)on the v-t graph slope of v-t graph  area of a-t graph a-t graph a-t graph

Refer to the following information for the next question.

Use the data analysis techniques shown in this lesson and the data given in the table below to determine if the object is experiencing uniformly accelerated motion.
 time(seconds) ending position (meters) 0 0 1 0.6 2 2.4 3 5.4 4 9.6 5 15.0 6 21.6

 a = Related Documents