Velocity vs. Time Graphs, Motion in One Dimension | Physics

Learning objectives

Section Key Terms

acceleration

Graphing Velocity as a Function of Time

Earlier, we examined graphs of position versus time. Now, we are going to build on that information as we look at graphs of velocity vs. time. Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity; we will discuss acceleration more in another chapter. These concepts are all very interrelated.

What can we learn about motion by looking at velocity vs. time graphs? Let’s return to our drive to school, and look at a graph of position versus time as shown in Figure_02_04_drive.

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Original figure description: A line graph titled Position versus Time is shown. The x-axis is labeled time in minutes and has a scale from zero to twenty in increments of one. The y-axis is labeled position, d, in kilometers and has a scale from zero to twelve in increments of one. The line intersects the following data points to form a line that slopes upward and then downward: zero, zero; two, one; four, two; ten, five; twelve, four; fourteen, three; twenty, zero.

A graph of position versus time for the drive to and from school is shown.

We assumed for our original calculation that your parent drove with a constant velocity to and from school. We now know that the car could not have gone from rest to a constant velocity without speeding up. So the actual graph would be curved on either end, but let’s make the same approximation as we did then, anyway.

Looking at this graph, and given what we learned, we can see that there are two distinct periods to the car’s motion—the way to school and the way back. The average velocity for the drive to school is 0.5 km/minute. We can see that the average velocity for the drive back is –0.5 km/minute. If we plot the data showing velocity versus time, we get another graph (Figure_02_04_velocity):

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Original figure description: A graph that shows time in minutes on the x-axis and velocity in kilometers per minute on the y axis. A horizontal line is show at a velocity of 0.5 that runs from 0 to 10 minutes. Another horizontal line is shown at a velocity of –0.5 that runs from 10 to 20 minutes.

Graph of velocity versus time for the drive to and from school.

We can learn a few things. First, we can derive a v versus t graph from a d versus t graph. Second, if we have a straight-line position–time graph that is positively or negatively sloped, it will yield a horizontal velocity graph. There are a few other interesting things to note. Just as we could use a position vs. time graph to determine velocity, we can use a velocity vs. time graph to determine position. We know that v = d/t. If we use a little algebra to re-arrange the equation, we see that d = v $\times$ t. In Figure_02_04_velocity, we have velocity on the y-axis and time along the x-axis. Let’s take just the first half of the motion. We get 0.5 km/minute $\times$ 10 minutes. The units for minutes cancel each other, and we get 5 km, which is the displacement for the trip to school. If we calculate the same for the return trip, we get –5 km. If we add them together, we see that the net displacement for the whole trip is 0 km, which it should be because we started and ended at the same place.

The area under a velocity curve represents the displacement. The velocity curve also tells us whether the car is speeding up. In our earlier example, we stated that the velocity was constant. So, the car is not speeding up. Graphically, you can see that the slope of these two lines is 0. This slope tells us that the car is not speeding up, or accelerating. We will do more with this information in a later chapter. For now, just remember that the area under the graph and the slope are the two important parts of the graph. Just like we could define a linear equation for the motion in a position vs. time graph, we can also define one for a velocity vs. time graph. As we said, the slope equals the acceleration, a. And in this graph, the y-intercept is v₀. Thus, $v = v_{0} + a t$ .

But what if the velocity is not constant? Let’s look back at our jet-car example. At the beginning of the motion, as the car is speeding up, we saw that its position is a curve, as shown in Figure_02_04_car.

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Original figure description: A line graph titled Jet Car Displacement is shown. The x-axis is labeled time, t, in seconds and has a scale from zero to forty on increments of ten. The y-axis is labeled displacement, x, in meters and has a scale from zero to three thousand five hundred in increments of five hundred. The following approximate data points are plotted, resulting in a line that curves upward: eight, two hundred fifty; ten, five hundred; fifteen, one thousand; twenty, one thousand five hundred; twenty-five, two thousand; thirty, three thousand. A right triangle is drawn at points eight, two hundred fifty; twelve, two hundred fifty, and twelve seven hundred fifty. The legs are labeled change in tp and change in dp. Point ten, five hundred is labeled P. Another right triangle is drawn at points twenty, one thousand five hundred; thirty, one thousand five hundred, and thirty, three thousand. The legs are labeled change in tq and change in dq. Point twenty-five, two thousand is labeled Q.

A graph is shown of the position of a jet-powered car during the time span when it is speeding up. The slope of a d vs. t graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point.

You do not have to do this, but you could, theoretically, take the instantaneous velocity at each point on this graph. If you did, you would get Figure_02_04_power, which is just a straight line with a positive slope.

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Original figure description: A line graph titled Jet Car Velocity is shown. The x-axis is labeled time, t, in seconds and has a scale from zero to forty on increments of ten. The y-axis is labeled velocity, v, in meters per second and has a scale from zero to one hundred eighty in increments of twenty. A straight line with a positive slope is plotted that intersects the following approximate points: zero, eighteen; ten, sixty (labeled P); fifteen, ninety; twenty, one hundred twenty; twenty five, one hundred fifty (labeled Q); thirty, one hundred sixty. Slope equals a is shown above the graphed lined.

The graph shows the velocity of a jet-powered car during the time span when it is speeding up.

Again, if we take the slope of the velocity vs. time graph, we get the acceleration, the rate of change of the velocity. And, if we take the area under the slope, we get back to the displacement.

Teacher Support

Return to the scenario of the drive to and from school. Re-draw the V-shaped position graph. Ask the students what the velocity is at different times on that graph. Students should then be able to see that the corresponding velocity graph is a horizontal line at 0.5km/minute and then a horizontal line at –0.5 km/minute. Then draw a few velocity graphs and see if they can get the corresponding position graph.

[OL][AL] Have students describe the relationship between the velocity and the position on these graphs. Ask—Can a velocity graph be used to find the position? Can a velocity graph be used to find anything else?

[AL] What is wrong with this graph? Ask students whether the velocity could actually be constant from rest or shift to negative so quickly. What would more realistic graphs look like? How inaccurate is it to ignore the non-constant portion of the motion?

[OL] Students should be able to see that if a position graph is a straight line, then the velocity graph will be a horizontal line. Also, the instantaneous velocity can be read off the velocity graph at any moment, but more steps are needed to calculate the average velocity.

[AL] Guide students in seeing that the area under the velocity curve is actually the position and the slope represents the rate of change of the velocity, just as the slope of the position line represents the rate of change of the position.

Solving Problems using Velocity–Time Graphs

Most velocity vs. time graphs will be straight lines. When this is the case, our calculations are fairly simple.

Using Velocity Graph to Calculate Some Stuff: Jet Car

Use this figure to (a) find the displacement of the jet car over the time shown (b) calculate the rate of change (acceleration) of the velocity. (c) give the instantaneous velocity at 5 s, and (d) calculate the average velocity over the interval shown.

Solution

1. Analyze the shape of the area to be calculated. In this case, the area is made up of a rectangle between 0 and 20 m/s stretching to 30 s. The area of a rectangle is length $\times$ width. Therefore, the area of this piece is 600 m.
2. Above that is a triangle whose base is 30 s and height is 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.
3. Add them together to get a net displacement of 2,700 m.
1. Take two points on the velocity line. Say, t = 5 s and t = 25 s. At t = 5 s, the value of v = 40 m/s.
  At t = 25 s, v = 140 m/s.
2. Find the slope. $\begin{matrix} a & = & \frac{Δ v}{Δ t} \\ = & \frac{100 m/s}{20 s} \\ = & 5 {m/s}^{2} \end{matrix}$
The instantaneous velocity at t = 5 s, as we found in part (b) is just 40 m/s.
1. Find the net displacement, which we found in part (a) was 2,700 m.
2. Find the total time which for this case is 30 s.
3. Divide 2,700 m/30 s = 90 m/s.

Discussion

The average velocity we calculated here makes sense if we look at the graph. 100m/s falls about halfway across the graph and since it is a straight line, we would expect about half the velocity to be above and half below.

Most of the velocity vs. time graphs we will look at will be simple to interpret. Occasionally, we will look at curved graphs of velocity vs. time. More often, these curved graphs occur when something is speeding up, often from rest. Let’s look back at a more realistic velocity vs. time graph of the jet car’s motion that takes this speeding up stage into account.

View original figure (external; image not reproduced in this text edition)

Original figure description: A line graph titled Jet Car Velocity is shown. The x-axis is labeled time, t, in seconds and has a scale from zero to eighty on increments of ten. The y-axis is labeled velocity, v, in meters per second and has a scale from one hundred sixty to two hundred sixty in increments of ten. The following approximate data points are plotted and connected to form a line that curves upward, and then becomes flat: zero, one hundred sixty five; ten, two hundred ten; twenty, two hundred thirty; thirty, two hundred forty; forty, two hundred forty-five; fifty, two hundred forty-five; sixty, two hundred forty-five; seventy, two hundred forty-five. A right triangle is drawn, with a hypotenuse that intersects points twenty, two hundred thirty and thirty, two hundred forty. The bottom leg of the triangle has ends at points zero, two hundred ten and fifty, two hundred ten and is labeled change in t. The upright leg has ends at points fifty, two hundred ten and fifty, two hundred sixty and is labeled change in v.

The graph shows a more accurate graph of the velocity of a jet-powered car during the time span when it is speeding up.

Using Curvy Velocity Graph to Calculate Some Stuff: Jet Car, Take Two

Use Figure_02_04_accurate to (a) find the approximate displacement of the jet car over the time shown, (b) calculate the instantaneous acceleration at t = 30 s, (c) find the instantaneous velocity at 30 s, and (d) calculate the approximate average velocity over the interval shown.

Solution

1. This problem is more complicated than the last example. To get a good estimate, we should probably break the curve into four sections. 0 → 10 s, 10 → 20 s, 20 → 40 s, and 40 → 70 s.
2. Calculate the bottom rectangle (common to all pieces). 165 m/s $\times$ 70 s = 11,550 m.
3. Estimate a triangle at the top, and calculate the area for each section. Section 1 contributes a triangle of width 10 s and height (210 – 165) m/s = 45 m/s so the added areas for Section 1 = (1/2)(45 m/s)(10 s) = 225 m; section 2 contributes a rectangle and a triangle with area = 100 m + 450 m = 550 m; section 3 = 150 m + 1,300 m = 1,450 m; section 4 = 2,550 m.
4. Add them together to get a net displacement of 16,325 m.
Using the tangent line given, we find that the slope is 1 m/s².
The instantaneous velocity at t = 30 s, is 240 m/s.
1. Find the net displacement, which we found in part (a), was 16,325 m.
2. Find the total time, which for this case is 70 s.
3. Divide $\frac{16,325 m}{70 s} \sim 233 m/s$

Discussion

This is a much more complicated process than the first problem. If we were to use these estimates to come up with the average velocity over just the first 30 s we would get about 191 m/s. By approximating that curve with a line, we get an average velocity of 202.5 m/s. Depending on our purposes and how precise an answer we need, sometimes calling a curve a straight line is a worthwhile approximation.

Practice Problems

View original figure (external; image not reproduced in this text edition)

Original figure description: The image shows a graph titled 'Velocity vs. Time' with velocity on the y-axis and time on the x-axis. The graphed line shows that velocity increases rapidly in 3 seconds to 3 m/s velocity but plateaus before losing velocity at 23 seconds.

Consider the velocity vs. time graph shown below of a person in an elevator. Suppose the elevator is initially at rest. It then speeds up for 3 seconds, maintains that velocity for 15 seconds, then slows down for 5 seconds until it stops. Find the instantaneous velocity at t = 10 s and t = 23 s.

Instantaneous velocity at t = 10 s and t = 23 s are 0 m/s and 0 m/s.
Instantaneous velocity at t = 10 s and t = 23 s are 0 m/s and 3 m/s.
Instantaneous velocity at t = 10 s and t = 23 s are 3 m/s and 0 m/s.
Instantaneous velocity at t = 10 s and t = 23 s are 3 m/s and 1.5 m/s.

View original figure (external; image not reproduced in this text edition)

Calculate the net displacement and the average velocity of the elevator over the time interval shown.

Net displacement is 45 m and average velocity is 2.10 m/s.
Net displacement is 45 m and average velocity is 2.28 m/s.
Net displacement is 57 m and average velocity is 2.66 m/s.
Net displacement is 57 m and average velocity is 2.48 m/s.

Check Your Understanding

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Section Summary

The slope of a velocity vs. time graph is the acceleration.
The area under a velocity vs. time curve is the displacement.
Average velocity can be found in a velocity vs. time graph by taking the weighted average of all the velocities.

Key Equations

Velocity	$v = v_{0} + a t$
Acceleration	$a = \frac{Δ v}{Δ t}$

Concept Items

A graph of velocity vs. time of a ship coming into a harbor is shown.

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Original figure description: The image shows a simple graph with velocity on the y-axis and time on the x-axis. The graphed line trends downward from left to right.

Describe the acceleration of the ship based on the graph.

The ship is moving in the forward direction at a steady rate. Then it accelerates in the forward direction and then decelerates.
The ship is moving in the forward direction at a steady rate. Then it turns around and starts decelerating, while traveling in the reverse direction. It then accelerates, but slowly.
The ship is moving in the forward direction at a steady rate. Then it decelerates in the forward direction, and then continues to slow down in the forward direction, but with more deceleration.
The ship is moving in the forward direction at a steady rate. Then it decelerates in the forward direction, and then continues to slow down in the forward direction, but with less deceleration.

Critical Thinking

Identify the time, or times, at which the instantaneous velocity is greatest, and the time, or times, at which it is negative. A sketch of velocity vs. time derived from the figure will aid in arriving at the correct answers.

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph plots time t on the x axis and position x on the y axis. The graph resembles a sine wave, beginning by going downward, then gradually sloping upward, then slightly downward, and more upward again. Points along the line are labeled a, b, c, d, e, f, g, h, I, j, k, and l.

The instantaneous velocity is greatest at f, and it is negative at d, h, I, j, and k.
The instantaneous velocity is greatest at e, and it is negative at a, b, and f.
The instantaneous velocity is greatest at f, and it is negative at d, h, I, j, and k
The instantaneous velocity is greatest at d, and it is negative at a, b, and f.

Problems

You are characterizing the motion of an object by measuring the location of the object at discrete moments in time. What is the minimum number of data points you would need to estimate the average acceleration of the object?

The correct answer is (c). A single point tells you only the object’s location at a particular time. Two points allow calculation of the average speed during that time interval, but do not provide information about the acceleration. Three data points are sufficient to provide information on the average acceleration, since they allow determination of average velocities for two different intervals in time.

Which option best describes the average acceleration from 40 to 70 s?

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Original figure description: A graph titled velocity vs. time plots time (s) on the x axis and velocity (m/s) on the y-axis. The line begins around 160 meters per second and approaches 250 meters per second as the time approaches 60 seconds.

It is negative and smaller in magnitude than the initial acceleration.
It is negative and larger in magnitude than the initial acceleration.
It is positive and smaller in magnitude than the initial acceleration.
It is positive and larger in magnitude than the initial acceleration.

The graph shows velocity vs. time.

View original figure (external; image not reproduced in this text edition)

Calculate the net displacement using seven different divisions. Calculate it again using two divisions: 0 → 40 s and 40 → 70 s . Compare. Using both, calculate the average velocity.

Displacement and average velocity using seven divisions are 14,312.5 m and 204.5 m/s while with two divisions are 15,500 m and 221.4 m/s respectively.
Displacement and average velocity using seven divisions are 15,500 m and 221.4 m/s while with two divisions are 14,312.5 m and 204.5 m/s respectively.
Displacement and average velocity using seven divisions are 15,500 m and 204.5 m/s while with two divisions are 14,312.5 m and 221.4 m/s respectively.
Displacement and average velocity using seven divisions are 14,312.5 m and 221.4 m/s while with two divisions are 15,500 m and 204.5 m/s respectively.

Performance Task

The National Mall in Washington, DC, is a national park containing most of the highly treasured memorials and museums of the United States. However, the National Mall also hosts many events and concerts. The map below shows the area for a benefit concert during which the president will speak. The concert stage is near the Lincoln Memorial. The seats and standing room for the crowd will stretch from the stage east to near the Washington Monument, as shown on the map. You are planning the logistics for the concert. Use the map scale to measure any distances needed to make the calculations below.

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Original figure description: A map drawing of the National Mall area of Washington, DC is shown. A red line is drawn on the map to show the presidential motorcade route from the White House at the top right of the map to a concert stage near the Lincoln Memorial at the bottom left of the map. Three points, one, two, and three, are plotted in the center of the map. Points one and two are directly across from each other. Point three is in between and slightly above points one and two.

The park has three new long-distance speakers. They would like to use these speakers to broadcast the concert audio to other parts of the National Mall. The speakers can project sound up to 0.35 miles away but they must be connected to one of the power supplies within the concert area. What is the minimum amount of wire needed for each speaker, in miles, in order to project the audio to the following areas? Assume that wire cannot be placed over buildings or any memorials.

Part A. The center of the Jefferson Memorial using power supply 1 (This will involve an elevated wire that can travel over water.)

Part B. The center of the Ellipse using power supply 3 (This wire cannot travel over water.)

Part C. The president’s motorcade will be traveling to the concert from the White House. To avoid concert traffic, the motorcade travels from the White House west down E Street and then turns south on 23rd Street to reach the Lincoln memorial. What minimum speed, in miles per hour to the nearest tenth, would the motorcade have to travel to make the trip in 5 minutes?

Part D. The president could also simply fly from the White House to the Lincoln Memorial using the presidential helicopter, Marine 1. How long would it take Marine 1, traveling slowly at 30 mph, to travel from directly above the White House landing zone (LZ) to directly above the Lincoln Memorial LZ? Disregard liftoff and landing times and report the travel time in minutes to the nearest minute.

Test Prep Multiple Choice

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Test Prep Short Answer

What would the velocity vs. time graph of the object whose position is shown in the graph look like?

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Original figure description: The image shows a graph titled 'Position vs Time' with position (m) on the y-axis and time (s) on the x-axis. The graphed data points show a trending curve upward from left to right.

It is a straight line with negative slope.
It is a straight line with positive slope.
It is a horizontal line at some negative value.
It is a horizontal line at some positive value.

Which statement correctly describes the object’s speed, as well as what a graph of acceleration vs. time would look like?

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Original figure description: A graph titled velocity vs. time plots time (s) on the x axis and velocity (m/s) on the y-axis. At time zero, the velocity is about 15 meters per second. The line extends at a constant rate and reaches a velocity of 170 at a time of 30 seconds.

The object is not speeding up, and the acceleration vs. time graph is a horizontal line at some negative value.
The object is not speeding up, and the acceleration vs. time graph is a horizontal line at some positive value.
The object is speeding up, and the acceleration vs. time graph is a horizontal line at some negative value.
The object is speeding up, and the acceleration vs. time graph is a horizontal line at some positive value.

Calculate that object’s net displacement over the time shown.

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540 m
2,520 m
2,790 m
5,040 m

What is the object’s average velocity?

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18 m/s
84 m/s
93 m/s
168 m/s

Test Prep Extended Response

Consider the motion of the object whose velocity is charted in the graph.

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Original figure description: A graph plots time t on the x axis and velocity v on the y axis. The line extends from the origin upward and to the right and plots points a, b, c, and d. At point d, the line begins to slope downward to point e. At point e, the line then begins to slope gradually upward through points f, g, and h, then drops off sharply through points I, j, k, and land on point l on the x axis again.

During which points is the object slowing down and speeding up?

It is slowing down between d and e. It is speeding up between a and d and e and h
It is slowing down between a and d and e and h. It is speeding up between d and e and then after i.
It is slowing down between d and e and then after h. It is speeding up between a and d and e and h.
It is slowing down between a and d and e and h. It is speeding up between d and e and then after i.

Divide the graph into approximate sections, and use those sections to graph the velocity vs. time of the object.

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph titled 'position vs time' with position (m) on the y-axis and time (s) on the x-axis. The data points trend upward for the first 5 seconds before moving downward by 10 seconds. After 10 seconds, the points gently curve upward with a max position of 25 m at 20 seconds.

Then calculate the acceleration during each section, and calculate the approximate average velocity.

Acceleration is zero and average velocity is 1.25 m/s.
Acceleration is constant with some positive value and average velocity is 1.25 m/s.
Acceleration is zero and average velocity is 0.25 m/s.
Acceleration is constant with some positive value and average velocity is 0.25 m/s.

Glossary

acceleration: the rate at which velocity changes

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