Contents

Acceleration

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)

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}}, \bar{a} = \frac{Δ v}{Δ t} = v_{f} - v_{0} t_{f} - t_{0}, size 12{ { bar  {a}}= {  {Δv}  over  {Δt} } = {  {v"" lSub { size 8{f} }  - v rSub { size 8{0} } }  over  {t rSub { size 8{f} }  - t rSub { size 8{0} } } } } {}

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where $\bar{a} \bar{a} size 12{ { bar {a}}} {}$ is average acceleration, $v v size 12{v} {}$ is velocity, and $t t size 12{t} {}$ is time. (The bar over the $a a size 12{a} {}$ means average acceleration.)

Because acceleration is velocity in m/s divided by time in s, the SI units for acceleration are ${m/s}^{2} {m/s}^{2} size 12{"m/s" rSup { size 8{2} } } {}$ , 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, $Δ v Δ v size 12{Dv} {}$ . 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 slows down, its acceleration is opposite to the direction of its motion. This is known as deceleration.

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)

Misconception Alert: Deceleration vs. Negative Acceleration

Deceleration always refers to acceleration in the direction opposite to the direction of the velocity. Deceleration always reduces speed. Negative acceleration, however, is acceleration in the negative direction in the chosen coordinate system. Negative acceleration may or may not be deceleration, and deceleration may or may not be considered negative acceleration. For example, consider Figure 2.14.

Four separate diagrams of cars moving. Diagram a: A car moving toward the right. A velocity vector arrow points toward the right. An acceleration vector arrow also points toward the right. Diagram b: A car moving toward the right in the positive x direction. A velocity vector arrow points toward the right. An acceleration vector arrow points toward the left. Diagram c: A car moving toward the left. A velocity vector arrow points toward the left. An acceleration vector arrow points toward the right. Diagram d: A car moving toward the left. A velocity vector arrow points toward the left. An acceleration vector arrow also points toward the left.

Figure 2.14 (a) This car is speeding up as it moves toward the right. It therefore has positive acceleration in our coordinate system. (b) This car is slowing down as it moves toward the right. Therefore, it has negative acceleration in our coordinate system, because its acceleration is toward the left. The car is also decelerating: the direction of its acceleration is opposite to its direction of motion. (c) This car is moving toward the left, but slowing down over time. Therefore, its acceleration is positive in our coordinate system because it is toward the right. However, the car is decelerating because its acceleration is opposite to its motion. (d) This car is speeding up as it moves toward the left. It has negative acceleration because it is accelerating toward the left. However, because its acceleration is in the same direction as its motion, it is speeding up (not decelerating).

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 2.15 (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 2.16

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

Solution

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

2. Find the change in velocity. Since the horse is going from zero to $- 15.0 m/s - 15.0 m/s size 12{ - "15" "." 0`"m/s"} {}$ , its change in velocity equals its final velocity: $Δ v = v_{f} = - 15 .0 m/s Δ v = v_{f} = - 15 .0 m/s$ .

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

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

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Discussion

The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of $8 .33 m {/s}^{2} 8 .33 m {/s}^{2}$ 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 $8 .33 m {/s}^{2} 8 .33 m {/s}^{2} size 12{8 "." "33"`"m/s" rSup { size 8{2} } } {}$ . 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 $a a$ , 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 2.17 shows graphs of instantaneous acceleration versus time for two very different motions. In Figure 2.17(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 $1 . 8 m {/s}^{2} 1 . 8 m {/s}^{2}$ ). In Figure 2.17(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 $+ 3 . 0 m {/s}^{2} + 3 . 0 m {/s}^{2}$ and $–2 . 0 m {/s}^{2} –2 . 0 m {/s}^{2}$ , respectively.

Figure 2.17 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 2.18. 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 2.18 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 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} x_{0}

x_{f} x_{f}

. Its displacement

Δ x Δ x

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

{x'}_{0} {x'}_{0}

{x'}_{f} {x'}_{f} size 12{ { {x}} sup { ' } rSub { size 8{f} } } {}

. Its displacement

Δ x' Δ x' size 12{Δx'} {}

- 1 .5 km - 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.)