3.1 Acceleration

Section Key Terms

Defining Acceleration

Throughout this chapter we will use the following terms: time, displacement, velocity, and acceleration. Recall that each of these terms has a designated variable and SI unit of measurement as follows:

• Time: t, measured in seconds (s)
• Displacement: Δd, measured in meters (m)
• Velocity: v, measured in meters per second (m/s)
• Acceleration: a, measured in meters per second per second (m/s², also called meters per second squared)
• Also note the following:
  • Δ means change in
  • The subscript 0 refers to an initial value; sometimes subscript i is instead used to refer to initial value.
  • The subscript f refers to final value
  • A bar over a symbol, such as $\overline{a}$, means average

Acceleration is the change in velocity divided by a period of time during which the change occurs. The SI units of velocity are m/s and the SI units for time are s, so the SI units for acceleration are m/s². Average acceleration is given by

Average acceleration is distinguished from instantaneous acceleration, which is acceleration at a specific instant in time. The magnitude of acceleration is often not constant over time. For example, runners in a race accelerate at a greater rate in the first second of a race than during the following seconds. You do not need to know all the instantaneous accelerations at all times to calculate average acceleration. All you need to know is the change in velocity (i.e., the final velocity minus the initial velocity) and the change in time (i.e., the final time minus the initial time), as shown in the formula. A component of the average acceleration can be positive, negative, or zero. A negative acceleration component is simply an acceleration in the negative direction along that axis. When the motion is in one dimension, we often simply refer to this as negative acceleration, and acceleration in the positive direction as positive acceleration.

Keep in mind that although acceleration points in the same direction as the change in velocity, it is not always in the direction of the velocity itself. When an object slows down, its acceleration is opposite to the direction of its velocity. In everyday language, this is called deceleration; but in physics, it is acceleration—whose direction happens to be opposite that of the velocity. For now, let us assume that motion to the right along the x-axis is positive and motion to the left is negative.

Figure 3.2 shows a car with positive acceleration in (a) and negative acceleration in (b). The arrows represent vectors showing both direction and magnitude of velocity and acceleration.

Figure 3.2 The car is speeding up in (a) and slowing down in (b).

Velocity and acceleration are both vector quantities. Recall that vectors have both magnitude and direction. An object traveling at a constant speed does accelerate if it changes direction. So, turning the steering wheel of a moving car makes the car accelerate because the velocity changes direction.

Virtual Physics

The Moving Man

With this animation, you can produce both variations of acceleration and velocity shown in Figure 3.2, plus a few more variations. Vary the velocity and acceleration by sliding the red and green markers along the scales. Keeping the velocity marker near zero will make the effect of acceleration more obvious. Try changing acceleration from positive to negative while the man is moving. We will come back to this animation and look at the Charts view when we study graphical representation of motion.

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Teacher Support

Teacher Support

Have students use a very low setting for velocity and acceleration because it is easier to see how the motion changes. Show students how setting velocity as positive and acceleration as negative creates the motion that resembles that of an object thrown into the air.

Grasp Check

Figure 3.3

Which part, (a) or (b), is represented when the velocity vector is on the positive side of the scale and the acceleration vector is set on the negative side of the scale? What does the car’s motion look like for the given scenario?

1. Part (a). The car is slowing down because the acceleration and the velocity vectors are acting in the opposite direction.
2. Part (a). The car is speeding up because the acceleration and the velocity vectors are acting in the same direction.
3. Part (b). The car is slowing down because the acceleration and velocity vectors are acting in the opposite directions.
4. Part (b). The car is speeding up because the acceleration and the velocity vectors are acting in the same direction.

Calculating Average Acceleration

Look back at the equation for average acceleration. You can see that the calculation of average acceleration involves three values: change in time, (Δt); change in velocity, (Δv); and acceleration (a).

Change in time is often stated as a time interval, and change in velocity can often be calculated by subtracting the initial velocity from the final velocity. Average acceleration is then simply change in velocity divided by change in time. Before you begin calculating, be sure that all distances and times have been converted to meters and seconds. Look at these examples of acceleration of a subway train.

Recall that for analysis of motion in only one dimension, we do not need to use vectors. Instead, we can treat displacement, velocity, and acceleration as scalars. Unlike magnitudes, scalars can be positive or negative, so the sign can serve to tell us the direction along the one-dimensional axis.

Worked Example

An Accelerating Subway Train

A subway train accelerates from rest to 30.0 km/h in 20.0 s. What is the average acceleration during that time interval?

Strategy

Start by making a simple sketch.

Figure 3.4

This problem involves four steps:

1. Convert to units of meters and seconds.
2. Determine the change in velocity.
3. Determine the change in time.
4. Use these values to calculate the average acceleration.

Solution

1. Identify the knowns. Be sure to read the problem for given information, which may not look like numbers. When the problem states that the train starts from rest, you can write down that the initial velocity is 0 m/s. Therefore, v₀ = 0; v_f = 30.0 km/h; and Δt = 20.0 s.

2. Convert the units.

   $$\frac{\text{30}\text{.0 km}}{\text{h}}\text{ × }\frac{{\text{10}}^{\text{3}}\text{m}}{\text{1 km}}\text{ × }\frac{\text{1 h}}{\text{3600 s}}\text{ = 8}\text{.333 }\frac{\text{m}}{\text{s}}$$

   3.1
3. Calculate change in velocity, $\Delta v=v_{\text{f}}\text{– }{v}_{\text{0}}\text{= 8}\text{.333 m/s – 0 = + 8}\text{.333 m/s,}$ where the plus sign means the change in velocity is to the right.

4. We know Δt, so all we have to do is insert the known values into the formula for average acceleration.

   $$\overline{a}=\frac{\Delta v}{\Delta t}=\frac{8.333\text{m/s}}{20.00\text{s}}=+0.417\frac{\text{m}}{{\text{s}}^{\text{2}}}$$

   3.2

Discussion

The plus sign in the answer means that acceleration is to the right. This is a reasonable conclusion because the train starts from rest and ends up with a velocity directed to the right (i.e., positive). So, acceleration is in the same direction as the change in velocity, as it should be.

Teacher Support

Teacher Support

Note that extra digits were carried along and rounding off to the correct number of significant figures, 3, was not done until the final answer was calculated.

Worked Example

An Accelerating Subway Train

Now, suppose that at the end of its trip, the train slows to a stop in 8.00 s from a speed of 30.0 km/h. What is its average acceleration during this time?

Strategy

Again, make a simple sketch.

Figure 3.5

In this case, the train is decelerating and its acceleration is negative because it is pointing to the left. As in the previous example, we must find the change in velocity and change in time, then solve for acceleration.

Solution

1. Identify the knowns: v₀ = 30.0 km/h; v_f = 0; and Δt = 8.00 s.
2. Convert the units. From the first problem, we know that 30.0 km/h = 8.333 m/s.
3. Calculate change in velocity, $\Delta v=v_{\text{f}}-v_0=0-8.333\text{m/s}=-8.333\text{m/s,}$ where the minus sign means that the change in velocity points to the left.

4. We know Δt = 8.00 s, so all we have to do is insert the known values into the equation for average acceleration.

   $$\overline{a}=\frac{\Delta v}{\Delta t}=\frac{-8.333\text{m/s}}{8.00\text{s}}=-1.04\frac{\text{m}}{{\text{s}}^{\text{2}}}$$

   3.3

Discussion

The minus sign indicates that acceleration is to the left. This is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would reduce the velocity. Again, acceleration is in the same direction as the change in velocity, which is negative in this case. This acceleration can be called a deceleration because it has a direction opposite to the velocity.

Teacher Support

Teacher Support

Help students see the relationship between the direction of the vector arrows and the plus and minus signs. Explain that one indication of the sign for acceleration is that it is in the direction opposite that of the velocity. Also point out that correctly identifying the initial and final speeds will result in the correct sign for acceleration.

Tips For Success

• It is easier to get plus and minus signs correct if you always assume that motion is away from zero and toward positive values on the x-axis. This way v always starts off being positive and points to the right. If speed is increasing, then acceleration is positive and also points to the right. If speed is decreasing, then acceleration is negative and points to the left.
• It is a good idea to carry two extra significant figures from step-to-step when making calculations. Do not round off with each step. When you arrive at the final answer, apply the rules of significant figures for the operations you carried out and round to the correct number of digits. Sometimes this will make your answer slightly more accurate.