Unit1 Part 5 Motion Graph Notes
Position-Time (AKA Displacement-Time) and Velocity-Time Graphs
Position-Time Graphs
Throughout the year you will be asked to graph the motion of any particular object. A problem might ask you to graph an object’s displacement over time or an object’s velocity over time. It’s important to remember that whenever you are graphing any motion in this class that time (t) is always on the x-axis; displacement or velocity will always be on the y-axis. Always.
+20m
+10m
d (m)
0m
0s10s20s30s
t (s)
The above displacement-time (AKA position-time) graph shows an object (it could be a person, a car, etc.) that is remaining at a certain position (not moving, standing still at the 10m mark) as time moves on. This particular graph shows constant position (not-changing position, zero displacement). If your position isn’t changing, this means you’re staying at the same place as time moves on. This would be similar to if you were running a race on the school’s track, got to the 10 m mark, and then stopped and didn’t move at all for the next 30 seconds. A common misperception students have when they read these types of graphs (constant, non-changing position), is that since the graph-line is actually continuing out in a horizontal direction that the object must be moving, too. When reading any graph it is very important to always look at BOTH the x- and y-axis to analyze the object’s motion. Again, in the above graph, the object is clearly staying at a single place/position while time moves forward. This is constant, not changing position; zero displacement.
+30m
+20m
+10m
d (m)
0m
0s10s20s30s
t (s)
The above d-t (displacement-time) graph shows non-constant, changing position. This graphs shows some displacement of the object. It shows an object beginning at the 0 meter mark (at t= 0 seconds) and then moving forward as time moves on. When you look at the graph you can see that the object is around the 10m mark 10s into the journey, at the 20m mark 20s into the journey, and at the 30 m mark 30s into the journey. You can tell that this is changing position as the object actually changes position. Any time you change position, you have displacement.
+50m
+40m
+30m
+20m
+10m
d (m)
0m
0s10s20s30s40s50s60s
t (s)
The above graph shows three different actions taking place over the 60 seconds: In the first 30 seconds (from t=0s to t=30s), the object’s position changes in the forward direction (it moves from 0m to +30m). This first motion represents changing (not constant) position (positive displacement) in the forward direction. In the next 20s (from t = 30 to 50s), the object has a constant position (not moving, zero displacment). In the last 10 seconds (from 50 to 60s) the object reverses direction and goes back to the 0m mark (from the +30m mark) in 10 seconds. This motion represents changing position in the reverse (negative) direction, which is negative displacement.
The Meaning Behind The Slope of Position-time (d-t) graphs
Remember from math that the slope of any line is simply determined by taking the “change in rise” over the “change in run”. This can also be thought of as the change in the y-axis (position in the d-t graphs) over the change in the x-axis (time in our graphs). You can put this into the following equation:
Slope = which is equal to (for our graphs) = which expands to =
If you look at the last equation, it should look familiar. Thinking back to the velocity notes, “change in position” over some time interval is equal to the velocity of any object. So, to pull it all together, the slope of a position-time (d-t) graph is equal to the average velocity ( v )of the object. This is super important, so put it in your memory right now.
Looking at the first graph from the “d-t graphs” section (re-drawn below), it is easy to see that the slope is zero (no slope). Remember that from math (horizontal lines have a slope of zero)????
+20m
+10m
d (m)
0m
0s10s20s30s
t (s)
However, if you want to calculate the actual value of the slope to verify that this is true, simply use “math” to figure out the value of the slope. To do this, take any two x-axis numbers and call them t1 (the time that comes first) and t2 (the time that comes second), and take the y-axis numbers that correspond to them and call them d1 and d2. Remember that d1 and t1 are the position and time that the object was at first, before the object was at d2 and t2. Then plug them into the slope equation (which is the equation for average velocity, v ).
Looking at the same graph immediately above, I’m choosing to call t1 = 0s. At that time, the object was at position 10m, so d1=10m. I’m choosing to call t2 = 30s. At that time, the object’s position was still 10m, so d2= 10m. Plugging those numbers into the equation to find slope (which is equal to velocity in the case of d-t graphs) shows
0 m/s
When the slope of any line is perfectly flat and horizontal, the slope of the line is zero. Mathematically we were able to determine that. Since the slope of any d-t graph is equal to the average velocity of the object, and the slope of the above graph is 0m/s, the velocity of the object in the above graph is 0m/s. This means that the object is moving zero meters every single second---i.e., the object isn’t moving.
+30m
+20m
+10m
d (m)
0m
0s10s20s30s
t (s)
Now find the value of the slope in the d-t graph immediately above. You can tell by looking at the graph that the slope is positive (remember that from math). This means that the velocity must also be positive. However, to calculate the actual numerical value of the slope (which is equal to the average velocity), you’d need to use the slope equation (change in rise over change in run). I’m going to use 0s as my first time and 30s as my second time. The corresponding positions for those two times are 0m and 30m.
+1 m/s
That positive “one meters/second” means that the object is traveling forward 1 meter every single second.
+50m
+40m
+30m
+20m
+10m
d (m)
0m
0s10s20s30s40s50s60s
t (s)
For the above graph there are three separate motions going on. In the first 30 seconds, the object is moving forward with some positive velocity. During the next 20 seconds, the object isn’t moving (the slope of the line is zero; the object’s position doesn’t change at all). During the last 10 seconds, the object is moving in the reverse direction from what it was during the first 30 seconds. It also is moving faster in the last part of its motion than it did in the beginning phase of its motion. You can tell this by noticing that it took 30s to cover 30m in the first phase, but it only took 10 seconds to cover the 30m in the last phase. You should also be able to tell just by looking at the graph that the slope of the line during the last 10 seconds is negative, but to get its actual numerical value you’d have to calculate its value with the “change in rise over change in run" equation. To calculate the velocities of the individual three motions, you’d have to use the slope equation for each of the three separate motions. Follow the directions below.
DO THIS: Pick any t1 and t2 and their corresponding d1 and d2 for each segment, and calculate the slope for each line segment/action.
ANSWERS: Line segment One:+1 m/s
Line Segment Two: 0 m/s
Line Segment Three: -3 m/s (again, the negative sign shows that it’s moving in the opposite direction of what we call “forward”)
Velocity-Time Graphs
Velocity-time graphs show how an object’s velocity changes over time. These graphs are very similar to
d-t graphs but they don’t show an object’s position, they show it’s velocity. If you know its velocity, you
can tell how fast it’s moving and in what direction.
+20m/s
+10m/s
v (m/s)
0m/s
0s10s20s30s
t (s)
The slope of the above line in the above graph (on the previous page) is zero (just by looking at it you can determine that). This means that the object’s velocity isn’t changing. However, it doesn’t mean the object isn’t moving. From looking at the graph, you can see that the velocity is a constant +10m/s. This means the object is moving forward (that’s what the positive sign of the velocity means) at a constant rate of 10 meters every single second.
+50m/s
+40m/s
+30m/s
+20m/s
+10m/s
v (m/s)
0m/s
0s10s20s30s40s50s60s
t (s)
Analyzing the motion of the above graph, there are (again) three different motions happening. During the first 30 seconds, the object’s velocity is increasing (changing) from 0m/s (rest, not moving) to +30m/s. This is like if you got in your car, started the engine, and began to speed up (accelerate). This changing velocity shows acceleration. The next 20 seconds (from t=30s to t=50s), the car is still moving, it’s just moving with a constant (non-changing) velocity of +30m/s. This would be like if you reached a certain speed in your car and remained at that constant speed for a period of time. This motion shows no acceleration. The last segment (t=50s to t=60s), the car’s velocity is negatively changing. For the graph above, this means that the object is slowing down---but it’s still moving forward while it’s doing that. This is just like you still moving forward when you’re in your car while you’re applying the brakes to slow down. The object doesn’t stop moving until its velocity is 0m/s.
The Meaning Behind The Slope of Velocity-time Graphs
Remember from our discussions on acceleration that acceleration refers to the change in velocity over
some time interval. In different words, any time an object’s velocity changes, it must be accelerating. So, looking at a v-t graph, anywhere on the graph that shows changing velocity, that also shows acceleration. If the slope of a “d-t graph” represents velocity, the slope of a “v-t graph” represents acceleration. To see why this is so, use the slope equation (change in rise over change in run) for a v-t graph. The “rise” on v-t graphs is velocity (units of m/s) rather than displacement (units of m) as in d-t graphs. So, your equation will be:
m/s/s
What the final above units mean is that the velocity is changing by a certain amount every single second.
+50m/s
+40m/s
+30m/s
+20m/s
+10m/s
v (m/s)
0m/s
0s10s20s30s40s50s60s
t (s)
Use the slope equation to find the slope for the three separate motions on the above v-t graph and you should get the following answers:
For t=0-30s a = +1m/s/s (positive acceleration; speeding up as it’s moving forward)
For t= 30-50sa= 0m/s/s (remaining at a constant speed; zero acceleration)
For t= 50-60sa = -3m/s/s (negative acceleration; slowing down as it’s moving forward)
The biggest thing to remember when analyzing any motion graphs is that d-t and v-t graphs represent
completely different things. A zero slope on a d-t graph means the object isn’t moving, but a zero slope on a v-t graph implies that the object is moving, just with a constant velocity (unless the horizontal line is along the 0m/s part of the v-t graph). Please keep in mind that whenever you are analyzing any motion graph, you need to really look at the x- and y-axis and really think about what the meaning is.
One last thing…..
The Area Under The Curve of Velocity-Time Graphs
+50m/s
+40m/s
+30m/s
+20m/s
+10m/s
v (m/s)
0m/s
0s10s20s30s40s50s60s
t (s)
Looking at the above v-t graph it is clear that the object is moving the whole time until it stops at 60 seconds (when its velocity becomes 0m/s). Since it’s moving, it must have covered some distance. You can determine the distance by analyzing the graph in a different way than we have in any of the other previous examples. Look at the first 30s. The object’s velocity changed in a constant way from 0m/s to 30m/s during that time. Looking at that portion of the graph a different way, the geometric shape of that part of the curve is a triangle with a base of 30s and a height of 30m/s. If you wanted to find the area of a triangle you would use the formula 1/2bh. Plugging our base and height values into that equation yields
Area = ½ (30s) (30m/s) 450m
The “seconds” cancel out in the above equation, yielding only distance/displacement units (meters). So, by taking the area under the curve on a v-t graph, you will find out the total distance the object traveled during that time.
For the next 20 seconds (from t=30s to t=50s), the shape of the curve is a rectangle (or a square). The height of the square is 30m/s and the base of the square is 20s. Plugging those values into the equation to find the area of a square (or rectangle)
Area= (30m/s)(20s)600m
Again, the “seconds” cancel out yielding only distance/displacement units. This means that during the next 20s (from 30 to 50s) the object moved a total of 600m.
The last 10 seconds the object is still moving forward, just slowing down. The shape of the curve is a triangle with a base of 10s and a height of 30m/s. Plugging these values into the equation for the area of triangle and you’ll get 150m. This means that during the last 10 seconds, the object moved 150m.
If you were asked to tell how far (distance) the object moved during the whole 60 seconds and all you were given was a v-t graph, you would need to find the total area under the curve. In order to do this, you would have to break it down into known geometric shapes that you can find the area of, find the individual areas of the “shapes”, and then add them all together. For the above graph, the total distance the object moved is
450m + 600m + 150m = 1200m
(Honors classes)
d
t
The above graph shows changing position (so it’s moving), but non-constant velocity. Anytime you have non-constant velocity, you’ve got acceleration. If the d-t graph was a straight line, it would show constant velocity (which is zero acceleration). But anytime a d-t graph shows non-constant velocity, it must represent accelerated motion. It shows that the object’s position changes by a non-constant amount during each time interval. This wouldn’t represent linear motion but exponential motion (y = x2). We’ll touch on this in class but you won’t get into it in depth until AP Physics (Calculus based).
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Unit 1 Part 5---Motion Graphs