Gregg Wolfe; Erika Gasper; John Stoke; Julie Kretchman; David Anderson; Nathan Czuba; Sudhi Oberoi; Liza Pujji; Irina Lyublinskaya; Douglas Ingram

An airplane flying very low to the ground, just above a beach full of onlookers, as it comes in for a landing.

Figure 2.12 A plane decelerates, or slows down, as it comes in for landing in St. Maarten. Its acceleration is opposite in direction to its velocity. (credit: Steve Conry, Flickr)

Learning Objectives

By the end of this section, you will be able to:

Define and distinguish between instantaneous acceleration and average acceleration.
Calculate acceleration given initial time, initial velocity, final time, and final velocity.

The information presented in this section supports the following AP® learning objectives and science practices:

3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2)
3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1)

In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater the acceleration, the greater the change in velocity over a given time. The formal definition of acceleration is consistent with these notions, but more inclusive.

Average Acceleration

Average Acceleration is the rate at which velocity changes,

\bar{a} = \frac{Δ v}{Δ t} = \frac{v_{f} - v_{0}}{t_{f} - t_{0}},

2.10

where $\bar{a}$ is average acceleration, $v$ is velocity, and $t$ is time. (The bar over the $a$ means average acceleration.)

Because acceleration is velocity in m/s divided by time in s, the SI units for acceleration are ${m/s}^{2}$ , meters per second squared or meters per second per second, which literally means by how many meters per second the velocity changes every second.

Recall that velocity is a vector—it has both magnitude and direction. This means that a change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.

Acceleration as a Vector

Acceleration is a vector in the same direction as the change in velocity, $Δ v$ . Since velocity is a vector, it can change either in magnitude or in direction. Acceleration is therefore a change in either speed or direction, or both.

Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction of motion. When an object's acceleration is in the same direction of its motion, the object will speed up. However, when an object's acceleration is opposite to the direction of its motion, the object will slow down. Speeding up and slowing down should not be confused with a positive and negative acceleration. The next two examples should help to make this distinction clear.

A subway train arriving at a station. A velocity vector arrow points along the track away from the train. An acceleration vector arrow points along the track toward the train.

Figure 2.13 A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of motion. (credit: Yusuke Kawasaki, Flickr)

Making Connections: Car Motion

The gray arrows are organized in 5 horizontal rows and labeled on the left edge as A, B, C, D, E. The left edge of the arrows on each row line up in a vertical line as well as the right edge of each row of arrows (the sum of all are the same length). The arrows in each of the rows are as follows: Row A's four arrows are pointing right: shortest, slightly longer, longer yet, longest. Row B's four arrows are pointing right: longest, slightly shorter, shorter yet, shortest (same direction but reverse order of sizes of A's row. Row C's arrows are pointing right: There are six arrows all the same as the shortest size of the arrows in row A and B (but the total length of all arrows and spaces is the same in all rows). Row D's four arrows are pointing left: longest, slightly shorter, shorter yet, shortest (same size but opposite direction of B's row). Row E's four arrows are pointing left: shortest, slightly longer, longer yet, longest (same size but opposite direction of A's row).

Figure 2.14 Above are arrows representing the motion of five cars (A–E). In all five cases, the positive direction should be considered to the right of the page.

Consider the acceleration and velocity of each car in terms of its direction of travel.

Figure 2.15 Car A is speeding up.

Because the positive direction is considered to the right of the paper, Car A is moving with a positive velocity. Because it is speeding up while moving with a positive velocity, its acceleration is also considered positive.

The diagram shows red car facing right in a road. Below the car is a B with four arrows pointing to the right: longest, slightly shorter, shorter yet, shortest.

Figure 2.16 Car B is slowing down.

Because the positive direction is considered to the right of the paper, Car B is also moving with a positive velocity. However, because it is slowing down while moving with a positive velocity, its acceleration is considered negative. (This can be viewed in a mathematical manner as well. If the car was originally moving with a velocity of +25 m/s, it is finishing with a speed less than that, like +5 m/s. Because the change in velocity is negative, the acceleration will be as well.)

The diagram shows red car facing right in a road. Below the car is a C with six arrows pointing to the right. The arrows are all the same as the shortest size of the arrows in row A and B.

Figure 2.17 Car C has a constant speed.

Because the positive direction is considered to the right of the paper, Car C is moving with a positive velocity. Because all arrows are of the same length, this car is not changing its speed. As a result, its change in velocity is zero, and its acceleration must be zero as well.

The diagram shows red car facing left in a road. Below the car is a D with four arrows pointing to the left: shortest, slightly longer, longer yet, longest (going from right to left – the same direction as the car).

Figure 2.18 Car D is speeding up in the opposite direction of Cars A, B, C.

Because the car is moving opposite to the positive direction, Car D is moving with a negative velocity. Because it is speeding up while moving in a negative direction, its acceleration is negative as well.

The diagram shows red car facing left in a road. Below the car is an E with four arrows pointing to the left: longest, slightly shorter, shorter yet, shortest (going from right to left – the same direction as the car).

Figure 2.19 Car E is slowing down in the same direction as Car D and opposite of Cars A, B, C.

Because it is moving opposite to the positive direction, Car E is moving with a negative velocity as well. However, because it is slowing down while moving in a negative direction, its acceleration is actually positive. As in example B, this may be more easily understood in a mathematical sense. The car is originally moving with a large negative velocity (−25 m/s) but slows to a final velocity that is less negative (−5 m/s). This change in velocity, from −25 m/s to −5 m/s, is actually a positive change ( $v_{f} - v_{i} = - 5 m/s - - 25 m/s$ of 20 m/s. Because the change in velocity is positive, the acceleration must also be positive.

Making Connection - Illustrative Example

The three graphs below are labeled A, B, and C. Each one represents the position of a moving object plotted against time.

The three graphs are labeled A, B, and C moving from left to right. The left, vertical axis on each graph is labeled Position. The bottom, horizontal axis is labeled Time. In graph A, the green curve line begins at the origin and starts horizontally with an increasing slope until the line is nearly vertical. In graph B, the green curve line begins at the origin and starts vertically with a decreasing slope until the line is nearly horizontal. In graph C, the green curve begins near the top of the Position axis and starts horizontally until it is nearly vertical at the end of the Time axis.

Figure 2.20 Three position and time graphs: A, B, and C.

As we did in the previous example, let's consider the acceleration and velocity of each object in terms of its direction of travel.

The green curve line begins at the origin and starts horizontally with an increasing slope until the line is nearly vertical.

Figure 2.21 Graph A of Position (y axis) vs. Time (x axis).

Object A is continually increasing its position in the positive direction. As a result, its velocity is considered positive.

Graph A above has a gray rectangle indicating about ½ of the horizontal Time and 1/5th of the vertical Position. The gray rectangle surrounds the green line on the bottom left of the graph and is longer than it is tall. A much larger green rectangle surrounds the last portion of the green curve on the right half of the graph. The width is only slightly less than the width of the gray rectangle and has about five times the height of the gray rectangle.

Figure 2.22 Breakdown of Graph A into two separate sections.

During the first portion of time (shaded grey) the position of the object does not change much, resulting in a small positive velocity. During a later portion of time (shaded green) the position of the object changes more, resulting in a larger positive velocity. Because this positive velocity is increasing over time, the acceleration of the object is considered positive.

In graph B, the green curve line begins at the origin and starts vertically with a decreasing slope until the line is nearly horizontal.

Figure 2.23 Graph B of Position (y axis) vs. Time (x axis).

As in case A, Object B is continually increasing its position in the positive direction. As a result, its velocity is considered positive.

Graph B above has a gray rectangle indicating about ½ of the horizontal Time and 4/5th of the vertical Position in the left half of the graph. The gray rectangle surrounds the green line and is taller than it is wide in the top half of the right side of the graph. A much smaller green rectangle surrounds the last portion of the green curve. The width is only slightly less than the width of the gray rectangle but has very little height (about 1/5th of the gray rectangle).

Figure 2.24 Breakdown of Graph B into two separate sections.

During the first portion of time (shaded grey) the position of the object changes a large amount, resulting in a large positive velocity. During a later portion of time (shaded green) the position of the object does not change as much, resulting in a smaller positive velocity. Because this positive velocity is decreasing over time, the acceleration of the object is considered negative.

In graph C, the green curve begins near the top of the Position axis and starts horizontally until it is nearly vertical at the end of the Time axis.

Figure 2.25 Graph C of Position (y axis) vs. Time (x axis).

Object C is continually decreasing its position in the positive direction. As a result, its velocity is considered negative.

Graph C above has a gray rectangle indicating about ½ of the horizontal Time and 1/5th of the vertical Position. The gray rectangle surrounds the green line and is much shorter than it is wide and starts at the top of the graph. A much larger green rectangle surrounds the last portion of the green curve on the right half of the graph. The width is only slightly less than the width of the gray rectangle and has about five times the height of the gray rectangle reaching to the Time axis.

Figure 2.26 Breakdown of Graph C into two separate sections.

During the first portion of time (shaded grey) the position of the object does not change a large amount, resulting in a small negative velocity. During a later portion of time (shaded green) the position of the object changes a much larger amount, resulting in a larger negative velocity. Because the velocity of the object is becoming more negative during the time period, the change in velocity is negative. As a result, the object experiences a negative acceleration.

Example 2.1

Calculating Acceleration: A Racehorse Leaves the Gate

A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration?

Figure 2.27 (credit: Jon Sullivan, PD Photo.org)

Strategy

First we draw a sketch and assign a coordinate system to the problem. This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity.

An acceleration vector arrow pointing west, in the negative x direction, labeled with a equals question mark. A velocity vector arrow also pointing toward the left, with initial velocity labeled as 0 and final velocity labeled as negative fifteen point 0 meters per second.

Figure 2.28

We can solve this problem by identifying $Δ v$ and $Δ t$ from the given information and then calculating the average acceleration directly from the equation $\bar{a} = \frac{Δ v}{Δ t} = \frac{v_{f} - v_{0}}{t_{f} - t_{0}}$ .

Solution

1. Identify the knowns. $v_{0} = 0$ , $v_{f} = - 15 .0 m/s$ (the minus sign indicates direction toward the west), $Δ t = 1 .80 s$ .

2. Find the change in velocity. Since the horse is going from zero to $- 15.0 m/s$ , its change in velocity equals its final velocity: $Δ v = v_{f} = - 15 .0 m/s$ .

3. Plug in the known values ( $Δ v$ and $Δ t$ ) and solve for the unknown $\bar{a}$ .

\bar{a} = \frac{Δ v}{Δ t} = \frac{- 15 .0 m/s}{1 .80 s} = - 8 .33 m {/s}^{2} .

2.11

Discussion

The minus sign for acceleration indicates that acceleration is toward the west. An acceleration of $8 .33 m {/s}^{2}$ due west means that the horse increases its velocity by 8.33 m/s due west each second, that is, 8.33 meters per second per second, which we write as $8 .33 m {/s}^{2}$ . This is truly an average acceleration, because the ride is not smooth. We shall see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.

Instantaneous Acceleration

Instantaneous acceleration $a$ , or the acceleration at a specific instant in time, is obtained by the same process as discussed for instantaneous velocity in Time, Velocity, and Speed—that is, by considering an infinitesimally small interval of time. How do we find instantaneous acceleration using only algebra? The answer is that we choose an average acceleration that is representative of the motion. Figure 2.29 shows graphs of instantaneous acceleration versus time for two very different motions. In Figure 2.29(a), the acceleration varies slightly and the average over the entire interval is nearly the same as the instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant acceleration equal to the average (in this case about $1 . 8 m {/s}^{2}$ ). In Figure 2.29(b), the acceleration varies drastically over time. In such situations it is best to consider smaller time intervals and choose an average acceleration for each. For example, we could consider motion over the time intervals from 0 to 1.0 s and from 1.0 to 3.0 s as separate motions with accelerations of $+ 3 . 0 m {/s}^{2}$ and $–2 . 0 m {/s}^{2}$ , respectively.

Figure 2.29 Graphs of instantaneous acceleration versus time for two different one-dimensional motions. (a) Here acceleration varies only slightly and is always in the same direction, since it is positive. The average over the interval is nearly the same as the acceleration at any given time. (b) Here the acceleration varies greatly, perhaps representing a package on a post office conveyor belt that is accelerated forward and backward as it bumps along. It is necessary to consider small time intervals (such as from 0 to 1.0 s) with constant or nearly constant acceleration in such a situation.

The next several examples consider the motion of the subway train shown in Figure 2.30. In (a) the shuttle moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems.

In part (a), a subway train moves from left to right from an initial position of x equals 4 point 7 kilometers to a final position of x equals 6 point 7 kilometers, with a displacement of 2 point 0 kilometers. In part (b), the train moves toward the left, from an initial position of 5 point 25 kilometers to a final position of 3 point 75 kilometers.

Figure 2.30 One-dimensional motion of a subway train considered in Example 2.2, Example 2.3, Example 2.4, Example 2.5, Example 2.6, and Example 2.7. Here we have chosen the

x

-axis so that + means to the right and

-

means to the left for displacements, velocities, and accelerations. (a) The subway train moves to the right from

x_{0}

to

x_{f}

. Its displacement

Δ x

is +2.0 km. (b) The train moves to the left from

{x'}_{0}

to

{x'}_{f}

. Its displacement

Δ x'

is

- 1 .5 km

. (Note that the prime symbol (′) is used simply to distinguish between displacement in the two different situations. The distances of travel and the size of the cars are on different scales to fit everything into the diagram.)

Example 2.2