3.2 Representing Acceleration with Equations and Graphs

Section Key Terms

How the Kinematic Equations are Related to Acceleration

We are studying concepts related to motion: time, displacement, velocity, and especially acceleration. We are only concerned with motion in one dimension. The kinematic equations that we will be using apply to conditions of constant acceleration, except where noted, and show how these concepts are related. Constant acceleration is acceleration that does not change over time. The first kinematic equation relates displacement d, average velocity $\overline{v}$, and time t.

The initial displacement $d_0$ is often 0, in which case the equation can be written as $\overline{v}=\frac{d}{t}$

This equation, which is the definition of average velocity and valid for both constant and non-constant acceleration, says that average velocity is displacement per unit time. We will express velocity in meters per second. If we graph displacement versus time, as in Figure 3.6, the slope will be the velocity. Whenever a rate, such as velocity, is represented graphically, time is usually taken to be the independent variable and is plotted along the x axis.

Figure 3.6 The slope of displacement versus time is velocity.

The second kinematic equation, another expression for average velocity $\overline{v},$ is simply the initial velocity plus the final velocity divided by two. This equation is only valid for constant acceleration.

Now we come to our main focus of this chapter; namely, the kinematic equations that describe motion with constant acceleration. In the third kinematic equation, acceleration is the rate at which velocity increases, so velocity at any point equals initial velocity plus acceleration multiplied by time

Note that this third kinematic equation does not have displacement in it. Therefore, if you do not know the displacement and are not trying to solve for a displacement, this equation might be a good one to use.

The third kinematic equation is also represented by the graph in Figure 3.7.

Figure 3.7 The slope of velocity versus time is acceleration.

The fourth kinematic equation shows how displacement is related to acceleration

When starting at the origin, $d_0=0$ and, when starting from rest, $v_0=0$, in which case the equation can be written as

This equation tells us that, for constant acceleration, the slope of a plot of 2d versus t² is acceleration, as shown in Figure 3.8.

Figure 3.8 When acceleration is constant, the slope of 2d versus t² gives the acceleration.

The fifth kinematic equation relates velocity, acceleration, and displacement

This equation is useful for when we do not know, or do not need to know, the time.

When starting from rest, the fifth equation simplifies to

According to this equation, a graph of velocity squared versus twice the displacement will have a slope equal to acceleration.

Figure 3.9

Note that, in reality, knowns and unknowns will vary. Sometimes you will want to rearrange a kinematic equation so that the knowns are the values on the axes and the unknown is the slope. Sometimes the intercept will not be at the origin (0,0). This will happen when v₀ or d₀ is not zero. This will be the case when the object of interest is already in motion, or the motion begins at some point other than at the origin of the coordinate system.

The kinematic equations are applicable when you have constant acceleration.

1. $d=d_0+\overline{v}t$, or $\overline{v}=\frac{d}{t}$ when d₀ = 0
2. $\overline{v}=\frac{v_0+v_{\text{f}}}{2}$
3. $v=v_0+at$, or $a=\frac{v}{t}$ when v₀ = 0
4. $d=d_0+v_{0}t+\frac{1}{2}a{t}^2$, or $a=\frac{2d}{t^2}$ when d₀ = 0 and v₀ = 0
5. $v^2=v_0^2+2a\left(d-d_0\right)$, or $a=\frac{v^2}{2d}$ when d₀ = 0 and v₀ = 0

Applying Kinematic Equations to Situations of Constant Acceleration

Problem-solving skills are essential to success in a science and life in general. The ability to apply broad physical principles, which are often represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Essential analytical skills will be developed by solving problems in this text and will be useful for understanding physics and science in general throughout your life.

Fun In Physics

Drag Racing

Figure 3.10 Smoke rises from the tires of a dragster at the beginning of a drag race. (Lt. Col. William Thurmond. Photo courtesy of U.S. Army.)

The object of the sport of drag racing is acceleration. Period! The races take place from a standing start on a straight one-quarter-mile (402 m) track. Usually two cars race side by side, and the winner is the driver who gets the car past the quarter-mile point first. At the finish line, the cars may be going more than 300 miles per hour (134 m/s). The driver then deploys a parachute to bring the car to a stop because it is unsafe to brake at such high speeds. The cars, called dragsters, are capable of accelerating at 26 m/s². By comparison, a typical sports car that is available to the general public can accelerate at about 5 m/s².

Several measurements are taken during each drag race:

• Reaction time is the time between the starting signal and when the front of the car crosses the starting line.
• Elapsed time is the time from when the vehicle crosses the starting line to when it crosses the finish line. The record is a little over 3 s.
• Speed is the average speed during the last 20 m before the finish line. The record is a little under 400 mph.

The video shows a race between two dragsters powered by jet engines. The actual race lasts about four seconds and is near the end of the video.

Teacher Support

Teacher Support

Explain the direction and magnitude of acceleration and velocity vectors before and after the braking chute is deployed. Explain that ¼ mile is 440 yards. If any students are on the track team, you might ask them to describe this distance. Compare record times for this track event to 4 seconds.

A dragster crosses the finish line with a velocity of $140\text{m/s}$. Assuming the vehicle maintained a constant acceleration from start to finish, what was its average velocity for the race?

1. $0\text{m/s}$
2. $35\text{m/s}$
3. $70\text{m/s}$
4. $140\text{m/s}$

Constant Acceleration

In many cases, acceleration is not uniform because the force acting on the accelerating object is not constant over time. A situation that gives constant acceleration is the acceleration of falling objects. When air resistance is not a factor, all objects near Earth’s surface fall with an acceleration of about 9.80 m/s². Although this value decreases slightly with increasing altitude, it may be assumed to be essentially constant. The value of 9.80 m/s² is labeled g and is referred to as acceleration due to gravity. Gravity is the force that causes nonsupported objects to accelerate downward—or, more precisely, toward the center of Earth. The magnitude of this force is called the weight of the object and is given by mg where m is the mass of the object (in kg). In places other than on Earth, such as the Moon or on other planets, g is not 9.80 m/s², but takes on other values. When using g for the acceleration a in a kinematic equation, it is usually given a negative sign because the acceleration due to gravity is downward.

Work In Physics

Effects of Rapid Acceleration

Figure 3.11 Astronauts train using G Force Simulators. (NASA)

When in a vehicle that accelerates rapidly, you experience a force on your entire body that accelerates your body. You feel this force in automobiles and slightly more on amusement park rides. For example, when you ride in a car that turns, the car applies a force on your body to make you accelerate in the direction in which the car is turning. If enough force is applied, you will accelerate at 9.80 m/s². This is the same as the acceleration due to gravity, so this force is called one G.

One G is the force required to accelerate an object at the acceleration due to gravity at Earth’s surface. Thus, one G for a paper cup is much less than one G for an elephant, because the elephant is much more massive and requires a greater force to make it accelerate at 9.80 m/s². For a person, a G of about 4 is so strong that his or her face will distort as the bones accelerate forward through the loose flesh. Other symptoms at extremely high Gs include changes in vision, loss of consciousness, and even death. The space shuttle produces about 3 Gs during takeoff and reentry. Some roller coasters and dragsters produce forces of around 4 Gs for their occupants. A fighter jet can produce up to 12 Gs during a sharp turn.

Astronauts and fighter pilots must undergo G-force training in simulators. The video shows the experience of several people undergoing this training.

People, such as astronauts, who work with G forces must also be trained to experience zero G—also called free fall or weightlessness—which can cause queasiness. NASA has an aircraft that allows it occupants to experience about 25 s of free fall. The aircraft is nicknamed the Vomit Comet.

Teacher Support

Teacher Support

Mention that, later in this course, students will encounter some interesting concepts related to gravity and acceleration when studying the general theory of relativity. In part, the theory is based on the idea that gravity cannot be distinguished from acceleration, either experientially or mathematically. When in an upward-bound elevator, can you really tell whether you are accelerating or whether gravity has suddenly become stronger?

Grasp Check

A common way to describe acceleration is to express it in multiples of g, Earth's gravitational acceleration. If a dragster accelerates at a rate of 39.2 m/s², how many g's does the driver experience?

1. 1.5 g
2. 4.0 g
3. 10.5 g
4. 24.5 g

Worked Example

Falling Objects

A person standing on the edge of a high cliff throws a rock straight up with an initial velocity v₀ of 13 m/s.

(a) Calculate the position and velocity of the rock at 1.00, 2.00, and 3.00 seconds after it is thrown. Ignore the effect of air resistance.

Strategy

Sketch the initial velocity and acceleration vectors and the axes.

Figure 3.12 Initial conditions for rock thrown straight up.

List the knowns: time t = 1.00 s, 2.00 s, and 3.00 s; initial velocity v₀ = 13 m/s; acceleration a = g = –9.80 m/s²; and position y₀ = 0 m

List the unknowns: y₁, y₂, and y₃; v₁, v₂, and v₃—where 1, 2, 3 refer to times 1.00 s, 2.00 s, and 3.00 s

Choose the equations.

$$d=d_0+v_{0}t+\frac{1}{2}a{t}^2\text{becomes}y=y_0+v_{0}t-\frac{1}{2}g{t}^2$$

3.11

$$v=v_0+at\text{becomes}v=v_0+-gt$$

3.12

These equations describe the unknowns in terms of knowns only.

Solution

$\begin{array}{c}{y}_1=0+\left(13.0\text{m/s}\right)\left(1.00\text{s}\right)+\frac{\left(-9.80{\text{m/s}}^2\right){\left(1.00\text{s}\right)}^2}{2}=8.10\text{m}\hfill \\ y_2=0+\left(13.0\text{m/s}\right)\left(2.00\text{s}\right)+\frac{\left(-9.80{\text{m/s}}^2\right){\left(2.00\text{s}\right)}^2}{2}=6.40\text{m}\hfill \\ y_3=0+\left(13.0\text{m/s}\right)\left(3.00\text{s}\right)+\frac{\left(-9.80{\text{m/s}}^2\right){\left(3.00\text{s}\right)}^2}{2}=-5.10\text{m}\hfill \\ v_1=13.0\text{m/s}+\left(-9.80{\text{m/s}}^2\right)\left(1.00\text{s}\right)=3.20\text{m/s}\hfill \\ v_2=13.0\text{m/s}+\left(-9.80{\text{m/s}}^2\right)\left(2.00\text{s}\right)=-6.60\text{m/s}\hfill \\ v_3=13.0\text{m/s}+\left(-9.80{\text{m/s}}^2\right)\left(3.00\text{s}\right)=-16.4\text{m/s}\hfill \end{array}$

Discussion

The first two positive values for y show that the rock is still above the edge of the cliff, and the third negative y value shows that it has passed the starting point and is below the cliff. Remember that we set up to be positive. Any position with a positive value is above the cliff, and any velocity with a positive value is an upward velocity. The first value for v is positive, so the rock is still on the way up. The second and third values for v are negative, so the rock is on its way down.

(b) Make graphs of position versus time, velocity versus time, and acceleration versus time. Use increments of 0.5 s in your graphs.

Strategy

Time is customarily plotted on the x-axis because it is the independent variable. Position versus time will not be linear, so calculate points for 0.50 s, 1.50 s, and 2.50 s. This will give a curve closer to the true, smooth shape.

Solution

The three graphs are shown in Figure 3.13.

Figure 3.13

Discussion

• y vs. t does not represent the two-dimensional parabolic path of a trajectory. The path of the rock is straight up and straight down. The slope of a line tangent to the curve at any point on the curve equals the velocity at that point—i.e., the instantaneous velocity.
• Note that the v vs. t line crosses the vertical axis at the initial velocity and crosses the horizontal axis at the time when the rock changes direction and begins to fall back to Earth. This plot is linear because acceleration is constant.
• The a vs. t plot also shows that acceleration is constant; that is, it does not change with time.

Teacher Support

Teacher Support

Prior to solving the problem, have students consider the following questions:

1. What is the goal?
2. What is known and unknown?
3. Which equation expresses the unknown in terms of the knowns?

After solving the problem, have students check that the answer is reasonable and has the correct units, sign, and significant figures.