Section Learning Objectives
By the end of this section, you will be able to do the following:
 Explain the meaning of slope and area in graphs of velocity vs. time
 Solve problems using speed vs. graphs. time
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teacher support
The learning objectives in this section will help your students master the following standards:
 (4)Science concepts. The student knows and applies the laws that govern movement in different situations. The student is expected to:
 (ONE) Generate and interpret graphs and charts describing different types of motion, including the use of realtime technology such as motion detectors or photogates.
Key Section Terms
acceleration
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teacher support
Ask students to use their knowledge of position graphs to construct graphs of velocity versus time. Alternatively, provide an example of a graph of velocity versus time and ask students what information can be derived from the graph. Ask: Is this the same information as a position versus time graph? How is information represented differently? Is there any new information on a speed vs. time?
Graph of speed versus time
Previously, we examinedgraphics depositioncontratempo🇧🇷 Now, let's build on this information as we look at the graphs ofSpeedwith respect to time Velocity is the rate of change ofdisplacement.Accelerationis the rate of change of velocity; We'll talk more about acceleration in another chapter. All these concepts are highly interrelated.
virtual physics
Maze
In this simulation, you will use a vector diagram to manipulate a ball to a certain location without hitting the wall. You can manipulate the ball directly with the position or by changing its speed. Explore how these factors change movement. If you want, you can put it ononefit too. This is acceleration, which measures the rate of change of velocity. We'll explore acceleration in more detail later, but it might be interesting to take a look here.
Click to see the content
If a person takes 3 steps and ends up exactly where they started, what must be true?

All three steps must have the same offset.

The displacement of the third step is greater than the displacement of the first two.

The average velocity must add up to zero.

Distance and average speed must add up to zero.
What can we learn about motion by looking at graphs of velocity versus time? Let's go back to our trip to school and see a graph of position versus time, as shown inFigure 2.15.
Figure 2.15 A position versus time graph for the trip to and from school is shown.
We assumed, for our original calculation, that your father drove at a constant speed to and from school. We now know that the car could not have gone from rest to a constant speed without accelerating. So the actual graph would be curved at each end, but let's do the same approximation as we did anyway.
tips for success
It is common in physics, especially in the early stages of learning, for certain things to becareless, as we see here. This is because it clarifies the concept or facilitates the calculation. Practicing physicists also use these types of shortcuts. It works because usually what it iscarelessis small enough not to significantly affect the response. In the example above, the time it takes for the car to accelerate and reach its cruising speed is very small compared to the total time traveled.
Looking at this graph, and considering what we've learned, we can see that there are two distinct periods in the car's motion: the way to school and the way back. The average speed for the commute to school is 0.5 km/minute. We can see that the average speed for the return trip is 0.5 km/minute. If we plot the data showing velocity versus time, we get another graph (Figure 2.16):
Figure 2.16 Graph of speed versus time for the trip to and from school.
We can learn a few things. First, we can derive avcontratgraph of adcontratgraphic. Second, if we have a straight line positiontime graph with a positive or negative slope, it will produce a horizontal velocity graph. There are some other interesting things to note. Just as we can use a position versus time graph to determine velocity, we can use a velocity versus time graph to determine position. We knowv=d/t🇧🇷 If we use a little algebra to rearrange the equation, we see thatd=v $\times $ t🇧🇷 InsideFigure 2.16, we have speed atyaxis and time along thexaxle. Let's take the first half of the movement. We get 0.5 km/minute$\times $10 minutes. the units forminutesthey cancel each other out, and we are left with 5 km, which is the displacement for the journey to school. If we calculate the same for the return trip, we get –5 km. If we add them up, we see that the net displacement for the entire trip is 0 km, which must be because we started and ended in the same place.
tips for success
You can treat units like you treat numbers, so km/km = 1 (or, say, cancels out). This is good because it can tell us whether or not we calculated everything in the correct units. For example, if we end up with m × s for velocity instead of m/s, we know something went wrong and we need to check our math. This process is called dimensional analysis, and it's one of the best ways to test whether your math makes sense in physics.
The area under a velocity curve represents displacement. The velocity curve also tells us whether the car is accelerating. In our previous example, we established that the velocity was constant. Therefore, the car is not accelerating. Graphically, you can see that the slope of these two lines is 0. This slope tells us that the car is not accelerating or is accelerating. We'll do more with this information in a later chapter. For now, remember that the area under the graph and the slope are the two important parts of the graph. Just as we can define a linear equation for motion on a position versus time graph, we can also define one for a velocity versus time graph. As we said, slope equals acceleration,one🇧🇷 And in this graph, theyintercept isv_{0}🇧🇷 This way,$v={v}_{0}+\mathrm{one}t$.
But what if the speed is not constant? Let's look at the example of our jet car. At the beginning of the movement, as the car accelerates, we see that its position is a curve, as shown in the figure.Figure 2.17.
Figure 2.17 A graph of a jet car's position is shown during the time period that it is accelerating. The slope of a d vs. t is the speed. This is shown in two points. The instantaneous velocity at any point is the slope of the tangent at that point.
You don't have to do this, but you could theoretically get the instantaneous velocity at each point on this graph. If you did, would youFigure 2.18, which is simply a straight line with a positive slope.
Figure 2.18 The graph shows the velocity of a jetpowered car during the time period that it accelerates.
Again, if we take the slope of the graph of velocity versus time, we get the acceleration, the rate of change of velocity. And if we take the area under the slope, we're back to displacement.
teacher support
teacher support
teacher demonstration
Return to the scenario of going to and from school. Redraw the graph of position in the shape of a V. Ask students what the velocity is at different times on this graph. Students should be able to see that the corresponding velocity graph is a horizontal line at 0.5 km/minute and then a horizontal line at –0.5 km/minute. Then draw some velocity graphs and see if you can get the corresponding position graph.
[OL][ALABAMA]Ask students to describe the relationship between velocity and position on these graphs. Ask: Can a velocity graph be used to find position? Can a velocity graph be used to find something else?
[ALABAMA]What's wrong with this chart? Ask students if the velocity could be constant from rest or become negative so quickly. What would the most realistic graphics look like? How inaccurate is it to ignore the nonconstant part of the motion?
[OL]Students should be able to see that if a graph of position is a straight line, then a graph of velocity is a horizontal line. Also, the instantaneous speed can be read from the speed graph at any time, but more steps are needed to calculate the average speed.
[ALABAMA]Guide students to see that the area under the velocity curve is actually position and that the slope represents the rate of change of velocity, just as the slope of the line of position represents the rate of change of position.
Solve problems using velocitytime graphs
Most speed vs. time will be straight lines. When this is the case, our calculations are quite simple.
worked example
Using Velocity Graph to Calculate Some Things: Jet Car
Use this figure to (a) find the displacement of the jet car during the time shown (b) calculate the rate of change (acceleration) of the velocity. (c) give the instantaneous velocity in 5 s, and (d) find the average velocity over the interval shown.
Strategy
 Displacement is obtained by finding the area under the line on the graph of velocity versus time.
 Acceleration is obtained by finding the slope of the velocity graph.
 Instantaneous speed can only be read from the graph.
 To find the average speed, remember that${v}_{\text{average}}=\frac{Dd}{Dt}=\frac{{d}_{\text{F}}{d}_{0}}{{t}_{\text{F}}{t}_{0}}$
Solution

 Analyze the shape of the area to be calculated. In this case, the area is formed by a rectangle between 0 and 20 m/s that extends to 30 s. the area of a rectangle is the length$\times $width. Therefore, the area of this piece is 600 m.
 Above is a triangle whose base measures 30 s and its height 140 m/s. the area of a triangle is 0.5$\times $length$\times $width. The area of this piece, therefore, is 2,100 m.
 Add them together to get a net displacement of 2700 m.

 Take two points on the speed line. To say,t= 5s et= 25 sec. ONEt= 5 s, the value ofv= 40m/s.
ONEt= 25 seconds,v= 140 m/s.  Find the slope.$\begin{array}{ccc}\hfill \mathrm{one}& =& \frac{Dv}{Dt}\hfill \\ & =& \frac{100\phantom{\rule{0ex}{0ex}}\text{millisecond}}{20\phantom{\rule{0ex}{0ex}}\text{s}}\hfill \\ & =& 5{\text{millisecond}}^{2}\hfill \end{array}$
 Take two points on the speed line. To say,t= 5s et= 25 sec. ONEt= 5 s, the value ofv= 40m/s.
 The instantaneous speed att= 5 s, as we discovered in part (b) is just 40 m/s.
 Find the net displacement, which we found in part (a) was 2700 m.
 Find the total time which for this case is 30 s.
 Divida 2700 m/30 s = 90 m/s.
Discussion
The average speed we calculated here makes sense if we look at the graph. 100 m/s falls roughly in the middle of the graph, and since it's a straight line, we expect about half the speed to be up and half down.
teacher support
teacher support
Solved quantities are slightly different for different types of graphs, but students should begin to see that the process of analyzing or breaking any of these graphs is similar. Ask: Where are the turning points in the movement? When does the object move? What does a curve in the graph mean? In addition, students should begin to have an intuitive understanding of the relationship between position and velocity graphs.
tips for success
You can have negative position, velocity, and acceleration on a graph that describes the way the object is moving. You should never see a graph with negative time on an axis. Because?
Most of the velocity vs. time graphs we'll see will be easy to interpret. Occasionally we will see graphs of velocity versus time. Most of the time, these curved graphs occur when something is accelerating, usually from rest. Let's look back at a more realistic speed versus time graph of the motion of the jet car that takes thisspeeding upstep under consideration.
Figure 2.19 The graph shows a more accurate graph of the velocity of a jetpowered car during the time interval that it is accelerating.
worked example
Using the Curvy Velocity Graph to Calculate a Few Things: Jet Car, Take Two
To useFigure 2.19To (a) find the approximate displacement of the jet car during the time shown, (b) calculate the instantaneous acceleration int= 30 s, (c) find the instantaneous velocity in 30 s, and (d) find the approximate average velocity over the interval shown.
Strategy
 Since this graph is an undefined curve, we have to estimate shapes at smaller intervals to find the areas.
 As when working with a curved displacement graph, we will need to take a tangent line at the instant of interest and use it to calculate the instantaneous acceleration.
 The instantaneous speed can still be read from the graph.
 We'll find the average velocity the same way we did in the previous example.
Solution

 This problem is more complicated than the last example. To get a good estimate, we should probably split the curve into four sections. 0 → 10 sec, 10 → 20 sec, 20 → 40 sec, and 40 → 70 sec.
 Calculate the bottom rectangle (common to all parts). 165m/s$\times $70 s = 11 550 m.
 Estimate a triangle at the top and find the area of each section. Section 1 = 225 m; section 2 = 100 m + 450 m = 550 m; stretch 3 = 150 m + 1,300 m = 1,450 m; stretch 4 = 2,550 m.
 Add them together to get a net displacement of 16,325 m.
 Using the given tangent line, we find that the slope is 1 m/s^{2}.
 The instantaneous speed att= 30 s, that's 240 m/s.
 Find the net displacement, which we found in part (a), was 16.325 m.
 Calculate the total time, which for this case is 70 s.
 Share$\frac{16.325\phantom{\rule{0ex}{0ex}}\text{metro}}{70\phantom{\rule{0ex}{0ex}}\text{s}}\sim 233\phantom{\rule{0ex}{0ex}}\text{millisecond}$
Discussion
This is a much more complicated process than the first problem. If we were to use these estimates to get the average velocity over just the first 30 s, we would get around 191 m/s. Approximating this curve with a line, we get an average velocity of 202.5 m/s. Depending on our purposes and how precise the answer we need, sometimes calling a curve a straight line is a worthwhile approximation.
teacher support
teacher support
Finding the tangent line can be a challenging concept for high school students and they need to understand it theoretically. If you've drawn a regular curve inside the curve at the point you're interested in, you can draw a radius of that curve. The tangent line would be the line perpendicular to that radius.
[OL]Ask students to compare this problem with the previous one. Ask: What's the difference? When would you care about the most accurate motion picture? And when would it really not matter? Why would you want to look at a less accurate representation of motion?
practice problems
20.
Figure 2.20
Consider the graph of velocity versus time below for a person in an elevator. Assume that the elevator is initially at rest. It then accelerates for 3 seconds, maintains that speed for 15 seconds, and slows down for 5 seconds until it stops. Find the instantaneous speed att= 10s et= 23 sec.
 instantaneous speed att= 10s et= 23 s filho 0 m/s y 0 m/s.
 instantaneous speed att= 10s et= 23 s filho 0 m/s y 3 m/s.
 instantaneous speed att= 10s et= 23 s filho 3 m/s y 0 m/s.
 instantaneous speed att= 10s et= 23 s son 3 m/s y 1,5 m/s.
21.
Figure 2.21
Calculate the net displacement and average speed of the elevator during the time interval shown.
 The net displacement is 45 m and the average velocity is 2.10 m/s.
 The net displacement is 45 m and the average speed is 2.28 m/s.
 The net displacement is 57 m and the average velocity is 2.66 m/s.
 The net displacement is 57 m and the average velocity is 2.48 m/s.
instant lab
Graphic motion, take two
In this activity, you will graph the speed of a moving ball against time.
 your graph from the previous Graphing Motion Snap Lab!
 1 sheet of graph paper
 1 pencil
Process
 Grab your graph from the previous Graphing Motion Snap Lab! and use it to create a graph of speed vs. time.
 Use your graph to calculate displacement.
22.
Describe the graph and explain what it means in terms of velocity and acceleration.

The graph shows a horizontal line indicating that the ball was moving at constant speed, i.e. it was not accelerating.

The graph shows a horizontal line indicating that the ball was moving with constant velocity, that is, it was accelerating.

The graph shows a horizontal line indicating that the ball was moving with variable velocity, i.e. it was not accelerating.

The graph shows a horizontal line indicating that the ball was moving with variable speed, that is, it was accelerating.
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teacher support
In this lab, students will use the displacement graph they drew in the last lab to create a velocity graph. If the rolling ball slowed down in the last lab, perhaps because the ramp was too low, the graph might not show a constant velocity.
check your understanding
23.
What information might you get from looking at a graph of velocity versus time?

acceleration

direction of movement

motion reference

shortest route
24.
How would you use a position versus time graph to construct a velocity versus time graph and vice versa?

The slope of a position versus time curve is used to construct a velocity versus time curve, and the slope of a velocity versus time curve is used to construct a position versus time curve.

The slope of a position versus time curve is used to construct a velocity versus time curve, and the area of a velocity versus time curve is used to construct a position versus time curve.

The area of a position versus time curve is used to construct a velocity versus time curve, and the slope of a velocity versus time curve is used to construct a position versus time curve.

The area of a position versus time curve is used to construct a velocity versus time curve, and the area of a velocity versus time curve is used to construct a position versus time curve.
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Use ocheck your understandingquestions to assess student performance against the section's learning objectives. If students are struggling with a specific goal, whatcheck your understandingwill help direct students to relevant content.