2.6: Acceleration

Learning Objectives

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

Figure \(\PageIndex{1}\): A plane slows down as it comes in for landing in St. Maarten. It is accelerating in a direction opposite 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.

Definition: AVERAGE ACCELERATION

Average acceleration is the rate at which velocity changes,

\[\bar{a}=\frac{\Delta v}{\Delta t}=\frac{v_{\mathrm{f}}-v_{0}}{t_{\mathrm{f}}-t_{0}}, \nonumber \]

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 \(\mathrm{m} / \mathrm{s}^{2}\), 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, \(\Delta v\). 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. If acceleration is in a direction opposite to the direction of motion, the object slows down.

Figure \(\PageIndex{2}\): A subway train in Sao Paulo, Brazil, slows down as it comes into a station. It is accelerating in a direction opposite to its direction of motion. (credit: Yusuke Kawasaki, Flickr)

Example \(\PageIndex{1}\): 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 \(\PageIndex{3}\)

Strategy

First we draw a sketch and assign a coordinate system to the problem. 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. (NOTE: you could have assigned west as positive as well, it is simply important to be aware of the direction as part of the answer because the average acceleration is a vector quantity.)

Figure \(\PageIndex{4}\)

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

Solution

1. Identify the knowns. \(v_{0}=0\), \(v_{\mathrm{f}}=-15.0 \mathrm{~m} / \mathrm{s}\) (the negative sign indicates direction toward the west), \(\Delta t=1.80 \mathrm{~s}\).

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: \(\Delta v=v_{\mathrm{f}}=-15.0 \mathrm{~m} / \mathrm{s}\).

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

\(\bar{a}=\frac{\Delta v}{\Delta t}=\frac{-15.0 \mathrm{~m} / \mathrm{s}}{1.80 \mathrm{~s}}=-8.33 \mathrm{~m} / \mathrm{s}^{2}\).

Discussion

The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of \(8.33 \mathrm{~m} / \mathrm{s}^{2}\) 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 \(8.33 \mathrm{~m} / \mathrm{s}^{2}\). 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—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 \(\PageIndex{5}\) shows graphs of instantaneous acceleration versus time for two very different motions. In Figure \(\PageIndex{5}\) (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 \(1.8 \mathrm{~m} / \mathrm{s}^{2}\)). In Figure \(\PageIndex{5}\) (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 \(+3.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(-2.0 \mathrm{~m} / \mathrm{s}^{2}\), respectively.

Figure \(\PageIndex{5}\): 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.

Exercise \(\PageIndex{1}\)

An airplane lands on a runway traveling east. Describe its acceleration.

Answer

If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also slowing down: its acceleration is opposite in direction to its velocity.

Section Summary

• Acceleration is the rate at which velocity changes. In symbols, average acceleration \(\bar{a}\) is

  \[\bar{a}=\frac{\Delta v}{\Delta t}=\frac{v_{\mathrm{f}}-v_{0}}{t_{\mathrm{f}}-t_{0}} \nonumber\]

• The SI unit for acceleration is \(\mathrm{m} / \mathrm{s}^{2}\).
• Acceleration is a vector, and thus has a both a magnitude and direction.
• Acceleration can be caused by either a change in the magnitude or the direction of the velocity.
• Instantaneous acceleration \(a\) is the acceleration at a specific instant in time.
• When an acceleration is in a direction opposite to that of the velocity of an object, the object slows down.

Glossary

acceleration

the rate of change in velocity; the change in velocity over time

average acceleration

the change in velocity divided by the time over which it changes

instantaneous acceleration

acceleration at a specific point in time

Contributors

Curated from resources found in Introduction to Physics published by OpenStax.