Acceleration

Saul Beceiro-Novo

Kinematics

11 Acceleration

Learning Objectives

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

An airplane flying very low to the ground, just above a beach full of onlookers, as it comes in for a landing. — Figure 10.1: A plane slows down (decelerates) as it prepares to land in St. Maarten. Its acceleration points in the opposite direction of its velocity. (credit: Steve Conry, Flickr)

In everyday language, the word “acceleration” usually means speeding up. And while this is certainly one case of acceleration, in physics, acceleration refers more broadly to any change in velocity—whether an object is speeding up, slowing down, or changing direction.

Average Acceleration

Average acceleration is defined as the change in velocity divided by the time interval over which that change occurs:

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

Here:

[latex]\bar{a}[/latex] is the average acceleration,
[latex]v_{f}[/latex] and [latex]v_{0}[/latex] are the final and initial velocities,
[latex]t_{f}[/latex] and [latex]t_{0}[/latex] are the final and initial times.

The SI unit of acceleration is meters per second squared [latex](m/s^2)[/latex], which literally means how much the velocity changes every second.

For example, if a plane slows down as it comes in for a landing, its velocity decreases over time. The acceleration in this case is in the opposite direction to the motion—this is a key idea in understanding deceleration, or negative acceleration.

Acceleration as a Vector

Like velocity, acceleration is a vector quantity, meaning it has both magnitude and direction. That direction is determined by the direction of the change in velocity

$(\Delta v)$ $(Δ v)$ . This means that an object can be accelerating even if its speed remains constant—if its direction is changing.

A classic example is a car turning a corner at a steady speed. Although the speed is unchanged, the direction of motion is changing, so the car is accelerating.

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 10.2: 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)

A common point of confusion arises when the acceleration of an object is in the opposite direction of its motion. This happens when an object slows down. In this case, the velocity and acceleration vectors point in opposite directions.

For instance, consider a subway train that decelerates as it approaches a station (see Figure 10.2). Even though the train is still moving forward, the acceleration is directed backward, against the motion.

This brings us to an important clarification:

Misconception Alert: Deceleration vs. Negative Acceleration

“Deceleration” refers to any situation in which the acceleration is in the opposite direction of the velocity—that is, the object is slowing down.

But “negative acceleration” refers only to the direction of acceleration in a coordinate system. Whether it causes an object to speed up or slow down depends on how the coordinate system is defined.

To make this clearer, look at the four cases illustrated in Figure 10.3:

(a) A car speeding up toward the right → positive acceleration
(b) A car slowing down toward the right → negative acceleration (also decelerating)
(c) A car slowing down toward the left → positive acceleration (but still decelerating)
(d) A car speeding up toward the left → negative acceleration, but not decelerating

As these cases show, the sign of acceleration depends on your chosen coordinate system. But deceleration always refers to a reduction in speed, regardless of direction.

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 10.3: (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 10.1: 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?

Two racehorses running toward the left. — Figure 10.4: (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 10.5

We can solve this problem by identifying [latex]\Delta v[/latex] and [latex]\Delta t[/latex] from the given information and then calculating the average acceleration directly from the equation [latex]\stackrel{-}{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}}[/latex].

Solution

Identify the knowns. [latex]{v}_{0}=0[/latex], [latex]{v}_{f}=-\text{15}\text{.0 m/s}[/latex] (the negative sign indicates direction toward the west), [latex]\Delta t=1\text{.80 s}[/latex].
Find the change in velocity. Since the horse is going from zero to [latex]-\text{15.0 m/s}[/latex], its change in velocity equals its final velocity: [latex]\Delta v={v}_{f}=-\text{15}\text{.0 m/s}[/latex].
Plug in the known values ([latex]\Delta v[/latex] and [latex]\Delta t[/latex]) and solve for the unknown [latex]\stackrel{-}{a}[/latex].
[latex]\stackrel{-}{a}=\frac{\Delta v}{\Delta t}=\phantom{\rule{0.15em}{0ex}}\frac{-\text{15}\text{.0 m/s}}{1\text{.80 s}}=-8\text{.33 m}{\text{/s}}^{2}.[/latex]

Discussion

The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of [latex]8\text{.33 m}{\text{/s}}^{2}[/latex] 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 [latex]8\text{.33 m}{\text{/s}}^{2}[/latex]. 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 [latex]a[/latex]—that is, the acceleration at a specific moment in time—is defined similarly to instantaneous velocity: it is the acceleration over an infinitesimally small time interval. In formal calculus-based physics, we define it as the derivative of velocity with respect to time. But in algebra-based physics, we can approximate it by using the average acceleration over very small time intervals when acceleration is approximately constant.

Choosing an Appropriate Approximation

Figure 10.6 shows acceleration-versus-time graphs for two very different types of motion:

In Figure 10.6(a), acceleration varies only slightly and remains positive throughout the time interval. In such a case, we can treat the motion as having a constant acceleration equal to the average:

[latex]\bar{a} = \frac{\Delta v}{\Delta t}[/latex]

For this particular case, we might take [latex]\bar{a} \approx 1.8 , \text{m/s}^2[/latex].
In contrast, Figure 10.6(b) shows motion in which acceleration changes drastically. This could represent, for example, a package being pushed forward and backward irregularly on a conveyor belt. In this case, it’s best to break the motion into smaller intervals and calculate average acceleration for each:

For the interval from [latex]0[/latex] to [latex]1.0 , \text{s}[/latex]:

[latex]\bar{a}_{0-1} = +3.0 , \text{m/s}^2[/latex]

For the interval from [latex]1.0[/latex] to [latex]3.0 , \text{s}[/latex]:

[latex]\bar{a}_{1-3} = -2.0 , \text{m/s}^2[/latex]

This piecewise approach gives a better representation of the actual motion.

Line graphs of instantaneous acceleration in meters per second per second versus time in seconds. The line on graph (a) shows slight variation above and below an average acceleration of about 1 point 8 meters per second per second. The line on graph (b) shows great variation over time, with instantaneous acceleration constant at 3 point 0 meters per second per second for 1 second, then dropping to negative 2 point 0 meters per second per second for the next 2 seconds, and then rising again, and so forth. — Figure 10.6: Instantaneous acceleration [latex]a[/latex] versus time [latex]t[/latex] for one-dimensional motion. (a) Acceleration is nearly constant and always positive. (b) Acceleration changes direction and magnitude significantly—e.g., a package jolting along a conveyor belt—requiring analysis over small [latex]\Delta t[/latex] intervals.

Subway Train Examples

Figure 10.7 shows two one-dimensional motions of a subway train, where we have chosen the [latex]x[/latex]-axis such that motion to the right is positive and motion to the left is negative.

In Figure 10.7(a), the train moves to the right, from an initial position [latex]x_0[/latex] to a final position [latex]x_f[/latex], giving a positive displacement:

[latex]\Delta x = x_f - x_0 = +2.0 , \text{km}[/latex]
In Figure 10.7(b), the train moves to the left, from an initial position [latex]x'_0[/latex] to a final position [latex]x'_f[/latex], resulting in a negative displacement:

[latex]\Delta x' = x'_f - x'_0 = -1.5 , \text{km}[/latex]

In both cases, acceleration is a vector quantity, so it depends not only on how much velocity changes but also on the direction of that change. Acceleration in the same direction as motion increases speed, while acceleration in the opposite direction reduces it (i.e., deceleration).

The next several examples consider the motion of the subway train shown in . 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 10.7: illustrates two examples of one-dimensional motion for a subway train, helping us understand how displacement, velocity, and acceleration behave along a chosen axis. In this case, the [latex]x[/latex]-axis is defined such that motion **to the right is positive** and motion **to the left is negative**. This convention applies consistently to displacements, velocities, and accelerations in the examples that follow.
(a) Motion to the Right
In **Figure 10.7(a)**, the subway train moves to the **right**, from an initial position [latex]x_0[/latex] to a final position [latex]x_f[/latex]. The displacement is:
[latex]\Delta x = x_f - x_0 = +2.0 , \text{km}[/latex]
Since the displacement is positive, we know the motion is in the positive [latex]x[/latex]-direction.
(b) Motion to the Left
In **Figure 10.7(b)**, the train moves in the opposite direction, to the **left**, starting from a position [latex]x'_0[/latex] and ending at [latex]x'_f[/latex]. Here, the displacement is:
[latex]\Delta x' = x'_f - x'_0 = -1.5 , \text{km}[/latex]
The negative sign in [latex]\Delta x'[/latex] indicates that the motion is in the negative [latex]x[/latex]-direction.
**Note**: The **prime symbol** (′) is used to distinguish the second scenario from the first. Additionally, the scales of motion and the size of the train cars in the figure are adjusted for visual clarity and may not reflect a common scale.