 GRAPHING ACCELERATION

Graphing Non-Uniform Motion and Using Tangent Lines to Find The Slope of a Curve at a Particular Moment in Time.

Now let's use some real world data courtesy of NASA.  Checkout the photo of the space shuttle taken from the orbiting International Space Station The space shuttle goes really, really fast. After launch, it goes from standing still to way faster than a speeding bullet in less than a minute.

However, the shuttle's velocity is definitely not constant.  Let's graph the following data: (NOTE:  Remember that the slope of a distance vs time graph is the velocity of the object)

Altitude (meters) Time (sec)
0 0
241 10
1,244 20
2,872 30
5,377 40
8,130 50
11,617 60
15,380 70
19,872 80
25,608 90
31,412 100
38,309 110
44,726 120

We are only graphing data in the positive Y direction, so we don't need any more labels than just plain meters. If you like, you can think of this as meters *up*!

We find that graphing that data shows not a straight line, but a curve.  What does that tell us? Recall that a straight line on a displacement vs time graph indicates constant velocity (constant speed in a single direction).  However, the space shuttle's velocity right after launch is changing very rapidly.  It is in fact, increasing every second which is a great example of acceleration which is very definitely non-constant velocity.

SLOPES

Remember, the slope of a displacement vs time graph is velocity.  Also, the slope of a line is defined to be how much the Y-axis values change in relation to the change in the X-axis values (recollect "rise over the run" from middle school!).

However, in the graph above, we aren't working with a straight line but instead we have a curve, so we can't talk about the slope of the curve because the curve is changing all the time.

So, what we do is to draw a straight line that is tangent or exactly matches what the curve is doing at a particular instant in time.  Once we have such a straight line, we can get the slope of that line.  And since we know that slope of a line on a distance vs time graph is velocity, we can determine the velocity of that object at that instant in time.  We call that instantaneous velocity.

Now, let's plot a few such points using that graph begining at time = 30 seconds (we'll assume 2 sig figs throughout).  Notice the red line in the following graph. As close as we can, we try to make it match what the curve is doing at just that point (time=30 seconds). Keep in mind that although we are making our best approximation, that the line isn't exact (which is fine since we are going for our best estimate).

Now let's find the slope of that red line.  We can take any two points on that line as reference and in fact we could make that line any length we want to make it easiest for ourselves.  Two points stand out as being easy references, (60, 10,000) and our current point, (30, 2500).  (Notice that we have to estimate the distance at time = 30 seconds, but again, this is an approximation so that is just fine).

Now that we have two points we can calculate the slope of that line, and since the slope of a line on a displacement vs time graph is velocity, we will have the velocity of the line tangent to the graph at t=30 seconds, or the instantaneous velocity at time = 30 seconds:

The change in Y is:  10,000 m - 2500 m = 7500 m

The change in X is: 60 sec - 30 sec = 30 sec

So the change in Y / Change in X = 250 m/sec.

Let's try again at time = 60 seconds Notice the green line is somewhat steeper (sorry it is kind of faint, my software kind of surprised me there).  Using the same method as above, we choose two points on the new line-- this time at

time = 70sec which has points (70 secs, 15,000 meters)

and

time = 40 sec ( which has points 40 secs, 5000 meters)

The change in Y is:  15,000 m - 5000 m = 10,000 m

The change in X is: 70 sec - 40 sec = 30 sec

So the change in Y / Change in X = 333 m/sec.

Let's try another at time = 90 seconds. Using the same method as above, we choose two points on the new (yellow) line-- this time at time = 80 (80, 20,000) and time = 100 (100, 30,000)

The change in Y is:  30,000 m - 20,000 m = 10,000 m

The change in X is: 100 sec - 80 sec = 20 sec

So the change in Y / Change in X = 500 m/sec.

Notice that the velocity of the shuttle is increasing, as we expected since the shuttle is in fact accelerating.

Let's try one last slope at time = 120 seconds. Using the same method as above, we choose two points on the new (black) line-- this time at time =

110 sec (110 sec, 40,000 meters)

and

time = 120 se (120, 45,000)

The change in Y is:  45,000 m - 40,000 m = 5,000 m

The change in X is: 120 sec - 110 sec = 10 sec

So the change in Y / Change in X = 500 m/sec.

So at this point, the shuttle has quit accelerating.

Now we can make a table of the velocities found above.

Velocity (m/s) Time (sec)
250 m/sec
30
333 m/sec
60
500 m/sec
90
500 m/sec
120

If we wanted to, we could graph this data on a velocity vs time graph. Thel SLOPE of that graph would give us the acceleration of the shuttle: 