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Motion Graphs Explanation, Review, and Examples introduction When trying to explain how things move, physicists don’t just use equations — they also use graphs! Motion graphs allow scientists to learn a lot about an object’s motion with just a quick glance. This article will cover the basics for interpreting motion graphs including different types of graphs, how to read them, and how they relate to each other. Interpreting motion graphs, such as position vs time graphs and velocity vs time graphs, requires knowledge of how to find slope. If you need a review or find yourself having trouble, this article should be able to help. What We Review Types of Motion Graphs Describing Motion with Position vs Time Graphs Position vs Time Graph for Multi-Stage Motion Position vs Time Graph for Passing an Observer Finding Distance and Displacement from a Position vs Time Graph Finding Velocity from a Position vs Time Graph Proving Velocity is the Slope of Position vs Time Graphs Describing Motion with Velocity vs Time Graphs Velocity vs Time Graph with a Change in Direction Finding Displacement from a Velocity vs Time Graph Proving Displacement is the Area under Velocity vs Time Graphs Finding Acceleration from a Velocity vs Time Graph Proving Acceleration is the Slope of Velocity vs Time Graphs Describing Motion with Acceleration vs Time Graphs Pairing Motion Graphs The Steps to Pairing Motion Graphs Conclusion Types of Motion Graphs There are three types of motion graphs that you will come across in the average high school physics course — position vs time graphs, velocity vs time graphs, and acceleration vs time graphs. An example of each one can be seen below. The position vs time graph (on the left) shows how far away something is relative to an observer. The velocity vs time graph (in the middle) shows you how quickly something is moving, again relative to an observer. Finally, the acceleration vs time graph (on the right) shows how quickly something is speeding up or slowing down, relative to an observer. Position (m) vs. Time (s) Velocity (m/s) vs. Time (s) Acceleration (m/s*2) vs. Time (8) 2 10 os io — = <== iB | peceeston (ne) 25 Tio (6) Time (@) Tino (0) Because all of these are visual representations of a movement, it is important to know your frame of reference. We learned in our introduction to kinematics that two people can observe the same event but describe it differently depending upon where they stand. If this or anything about the position, velocity, and/or acceleration is still a bit confusing, revisit our kinematics post and our acceleration post before moving on. Describing Motion with Position vs Time Graphs We typically start with position-time graphs when . . ‘ Position (m) vs. Time (s) learning how to interpret motion graphs — generally because they’re the easiest to try to picture. Let’s look at the position vs time graph from above. We see that our vertical axis is Position (in meters) and that our horizontal axis is Time (in seconds). This means we know how far Position (m) away an object has moved from our observer at any given time. This particular graph shows an ° 1 2 3 4 5 object moving steadily away from our observer. rae Position vs Time Graph for Passing an Observer This implies that the position could, potentially, go below the x-axis. Let’s look at an example combining these two points in practice. Position (m) vs. Time (s) 10 5 E 5 8 E 5 -10 Time (s) t=10s t=0s C ) x=-10m f x=10m This time, our car started to the right, and drove straight past our observer to the left. Att = 0 s, the car was 10 m to the right of our observer, so its position was x = 10 m.Asit passed the observer, its position was 2 = 0 matt = 5s. The car then ended its journey 10 m to our observers left at t = 10 s so that its final position was z = —10 m. Finding Distance and Displacement from a Position vs Time Graph Example 1: Constant Position vs Time Graph We'll continue working from the graph above as we have already pulled the important values from it. Because we have a simple, straight line we only need the values from the very beginning and very end of the car’s journey, which we already pulled out above: et; =0s *2;=10m eto=10s ez=-—10m Finding Distance From A Position-Time Graph As we learned from our introduction to kinematics lesson, we know that the equation for distance is: Formula for Distance dt = d, + do Example 2: Changing Position vs Time Graph Now that we know the basics of finding distance and displacement from a position vs time graph, let's get a bit more in-depth. We'll return to the graph about the car that moved forward, stopped, and then turned around and returned to its original position. The graph has been copied below for convenience. Position (m) vs. Time (s) 6 4 E € s g 2 oO 0 5 10 15 Time (s) How to Find Distance From A Position vs Time Graph Finding distance from these graphs can get a bit complex as you'll need to find several different values. If you'll notice, the slope of our graph changes regularly — the line seems to turn. Each segment with a unique slope requires our attention. So, we'llneed to look att = 0 sthrought = 5s,t = 5 sthrough t = 10 s,and t= 10sthrought = 15s. We'll want to look at the position value on the left and right of each side of those segments and find the absolute value of the delta between those values. These will serve as the d values that we will plug into our distance equation. edq=)m—Om\=5m ed=|5m—5m\=0m ed3=|Om—5m\|=5m We can now plug all of these values into our equation and solve for distance. dp =dj+dg9+d3=5m+0m+5m=10m How to Find Displacement From A Position vs Time Graph Finding displacement from a graph that changes how it’s moving is a bit easier than finding the distance. Because displacement only concerns the distance between the starting and ending positions of an object's motion, we only need to find the position at the rightmost point on the graph (t = 15 s) and the leftmost point on the graph (t = 0 s). The positions at these times will serve as our @ f and Zj values respectively. . vf = 0 m -2zj=O0m Now that we have these values, we can plug them into our displacement formula and solve: Ar=2;—-2;=0m—-0Om=0m Finding Velocity from a Position vs Time Graph Now that we know how to find distance and displacement from a position vs time graph, we can start finding another value — velocity. If you think about it, these distances and displacements that we're finding are occurring over some amount of time (as given by the graph) and all we really need to find velocity is displacement and time. So let’s start with a simple graph — the one of an object moving steadily away. {he dispiscemcntrarthe Position (m) vs. Time (s) movement depicted by this graph 25 would be Az = 25m 0 m = 25 mand because our time here moves fromt = O sto t = 5 s,we have achangein time of At = 5 s. This is enough information for us to solve for the velocity using the equation a5 Position (m) 5 we learned before: ° ,— 42 _m “ "= At 5s =5m/s We have a value that matches the velocity we solved for in both numerical value and physical units. We could have selected any two points on our graph and received the same result. The part of this that truly proves the slope of any position-time graph is the velocity is that the units of our slope work out to be m/s - the units for velocity. Example 1: Finding Velocity from a Position vs Time Graph Let's try finding our velocity again, but using the slope formula. We'll reuse the graph below that we saw earlier in this article. Position (m) vs. Time (s) 10 5 E zg % a o a 5 -10 ie) 2 4 6 8 10 Time (s) We'll want to begin by selecting the points we want to use. This graph is a straight line which means its slope never changes so it won't particularly matter what two points we choose. Since we have a point where our Z value is zero and a separate point where our y value is zero, we may as well use those to make the mathematics easier. So, the values we'll be using here are: yo=Om -y=—10m © m=5s ° 271=0s Now, all we need to do is set our velocity equal to our slope, plug in our values, and solve for our velocity: _—yw-y _Om-—10m -10m = a) zr — 2} 5s-—Os 5s a a Here, we get a negative velocity of u = —2 m/s. If we look at our graph, we see it has a negative slope, so we should have expected this negative velocity from the start. If you ever get a positive when you expected a negative or vice versa, check to make sure you plugged your values into your formula in the correct order. That simple mistake has thrown many scientists off course. Example 2: Finding Velocity with Changing Motion Being able to find the velocity of a simple, straight position vs time graph is all well and good, but there will be times when you'll have to split a graph apart. Let's revisit the graph below as an example of this. Position (m) vs. Time (s) 6 Position (m) Time (s) To find the velocity Velocity (m/s) vs. Time (s) 10 of an object at any given time here, we simply need to read the value from the 6 graph. There's no mathematics to do or formulas to use. Velocity (mis) So, for example, at t = 2 sthe velocity isu =4m/s because that is the Time (s) value we read off of the graph. Similarly, the velocity ist = 4 sisv = 8 m/s. The fact that those two values differ and that the slope here is positive tells us that the motion in this graph is an object moving away from an observer and getting faster - like a car leaving a stoplight. We can also show more complicated motions and dip below the x-axis. Velocity vs Time Graph with a Change in Direction Let's imagine a scenario for the graph to the Velocity (m/s) vs. Time (s) right. We see that the graph starts with the bl object's top velocity of v = 10 m/s and ® then seems to get lower. The object reaches avelocity of v = 0 m/satt = 2.5 s. While it may make sense to say that the object is > Vetocty ¢rts) now at the same point as the observer, we | L can’t actually infer that. All we can tell from “Ze here is that the object is momentarily at rest relative to the observer. The velocity then continues decreasingtov = —10m if $; implying that the object is now moving in the negative direction. A real-life scenario for this may be that you observe someone pulling into a long driveway, stopping briefly at the end, and then backing down it. Velocity vs Time Graph for Multi-Stage Motion Now, let’s return to our car from before that moved in the positive direction. stopped, and then came back. Since the slope in each segment of the position graph was constant, we assumed that the car’s movements had a constant velocity and that it had zero velocity when it stopped. The velocity-time graph for this motion would look a bit like this: Velocity (m/s) vs. Time (s) 2 Velocaty (m/s) he Time (s) You can see that the velocity remains a constant v = 1 m/s while the car moves to the right, changestov = 0m / s while the car stops, and then becomes v = 1 m/swhile the car moves back to the left. Finding Displacement from a Velocity vs Time Graph Much like how we could find a velocity from a position-time graph, we can find displacement from a velocity-time graph. This process will be a bit different. Instead of finding the slope of the velocity graph, we will be finding the area under the velocity graph. This may sound counterintuitive, but we can prove that it works by checking our units. Let’s say that an object moves at 5 m/s for 10 s. The velocity-time graph for this motion would look like this: It’s worth noting here that the units along each axis were also included for the base and height of the rectangle. The equation for the area under the curve is the one you would use to find the area of arectangle, A = bh. So, let’s pull down our values and solve our equation: A=bh=(10s)-(5m/s) =50m As aresult, we obtained the same numerical value of 50, but more to the point we obtained the correct units. The area under the curve of a velocity graph will always be a displacement. Let’s look at a couple of more examples. If you're uncertain about your ability to remember the equations for the area of a rectangle or triangle, it may be worth writing them in your notes or referencing a formula sheet such as this one. Example 1: Finding Displacement for Multiple Velocities The graph above was pretty simple, so let's look at some more complex motion graphs. We can return to the velocity-time graph for our car that moved to the right, paused, then moved back to the left. Velocity (m/s) vs. Time (s) 2 Time (s) We already know that our displacement for this motion is 0 m. Let’s start by sectioning off our graph here into shapes we cam find the area of. Again, we're looking for the area between the line of the graph and the x-axis. Velocity (m/s) vs. Time (s) It seems strange to havea . negative value for the height of a shape as you've likely been told that area should always be a positive value. We'll see why having a negative height when the graph is below the x-axis is both allowed and important. Now that Velocty Gms) we have all of our rectangles we can start finding their area. Let's begin with the rectangle farthest to the left. «© b=5s -h=1m/s Aleft = bh = (5s)-(1m/s)=5m Now we can solve for the area of our middle rectangle. This may seem like a trick question as it is, essentially, just a flat line, but we'll still want to include it. «b=5s -h=0m/s Amidate = bh = (5s) -(0m/s) = 0m Finally. let’s find the area of the rectangle on the right. This has a negative value for its height so it should also have a negative area, strange as that may seem. We can create a rectangle on the right where the velocity is constant, but the area where it’s increasing will not look like a rectangle at all. Instead, this is where we'll have to create a section that is a triangle. We now have two velocity (m/s) vs. time (s) separate shapes. Much e like when we had three 10 separate rectangles, we'll find the area of each shape individually and then add those two areas together to 2 velocity (ms) - ° find the overall Fy displacement for this motion. Let’s start with the triangle. *b=5s «h=10m/s 1 1 ATriangle = bh = (5 s) . (10 m/s) =25m Now, we can find the area of the rectangle portion. «b=5s « h=10m/s ARectangle = bh = (5s) - (10 m/s) = 50m Finally, we'll add these two values to find our total displacement: Az = ATotal = ATriangle + ARectangle = 25m + 50m = 75m Finding Acceleration from a Velocity vs Time Graph At this point, it may not shock you to learn that the slope of a velocity-time graph can tell us just as much as the area under its curve. Instead of displacement, though, the slope of a velocity graph will tell us an object’s acceleration. Let’s consider the graph. velocity (m/s) vs. time (s) 12 10 velocity (m/s) time (s) The velocity of the object being shown in this graph is steadily increasing by 2 m / s every 1 s. With that information, we can prove thata = Av/At = (2 m/s)/(1s) =2 m/s”. Now that we know what our acceleration should be, let's try to find it by finding the slope of the velocity time graph. Proving Acceleration is the Slope of Velocity vs Time Graphs If you remember from earlier, the slope of any given straight line can be found with the equation: Slope Formula —_ oe mQ—-21 m 
