### Speed

There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we will look at time, speed, and velocity to expand our understanding of motion.

A description of how fast or slow an object moves is its speed. Speed is the rate at which an object changes its location. Like distance, speed is a scalar because it has a magnitude but not a direction. Because speed is a rate, it depends on the time interval of motion. You can calculate the elapsed time or the change in time, $\Delta t$, of motion as the difference between the ending time and the beginning time

$$\Delta t={t}_{\text{f}}-{t}_{\text{0}}.$$

The SI unit of time is the second (s), and the SI unit of speed is meters per second (m/s), but sometimes kilometers per hour (km/h), miles per hour (mph) or other units of speed are used.

When you describe an object's speed, you often describe the average over a time period. Average speed, *v*_{avg}, is the distance traveled divided by the time during which the motion occurs.

$${v}_{\text{avg}}=\frac{\text{distance}}{\text{time}}$$

You can, of course, rearrange the equation to solve for either distance or time

$$\text{time=}\frac{\text{distance}}{{v}_{\text{avg}}}\text{.}$$

$$\text{distance=}{v}_{\text{avg}}\text{\xd7time}$$

Suppose, for example, a car travels 150 kilometers in 3.2 hours. Its average speed for the trip is

$$\begin{array}{ccc}\hfill {v}_{\text{avg}}& =& \frac{\text{distance}}{\text{time}}\hfill \\ & =& \frac{150\text{km}}{3.2\text{h}}\hfill \\ & =& 47\text{km/h.}\hfill \end{array}$$

A car's speed would likely increase and decrease many times over a 3.2 hour trip. Its speed at a specific instant in time, however, is its instantaneous speed. A car's speedometer describes its instantaneous speed.

### Worked Example

#### Calculating Average Speed

A marble rolls 5.2 m in 1.8 s. What was the marble's average speed?

### Strategy

We know the distance the marble travels, 5.2 m, and the time interval, 1.8 s. We can use these values in the average speed equation.

Solution

$${v}_{\text{avg}}=\frac{\text{distance}}{\text{time}}=\frac{5.2\text{m}}{1.8\text{s}}=2.9\text{m/s}$$

Discussion

Average speed is a scalar, so we do not include direction in the answer. We can check the reasonableness of the answer by estimating: 5 meters divided by 2 seconds is 2.5 m/s. Since 2.5 m/s is close to 2.9 m/s, the answer is reasonable. This is about the speed of a brisk walk, so it also makes sense.

#### Practice Problems

A pitcher throws a baseball from the pitcher’s mound to home plate in 0.46 s. The distance is 18.4 m. What was the average speed of the baseball?

- 40 m/s
- - 40 m/s
- 0.03 m/s
- 8.5 m/s

Cassie walked to her friend’s house with an average speed of 1.40 m/s. The distance between the houses is 205 m. How long did the trip take her?

- 146 s
- 0.01 s
- 2.50 min
- 287 s

#### Velocity

The vector version of speed is velocity. Velocity describes the speed and direction of an object. As with speed, it is useful to describe either the average velocity over a time period or the velocity at a specific moment. Average velocity is displacement divided by the time over which the displacement occurs.

$${v}_{\text{avg}}=\frac{\text{distance}}{\text{time}}=\frac{\Delta d}{\Delta t}=\frac{{d}_{\text{f}}-{d}_{0}}{{t}_{\text{f}}-{t}_{0}}$$

Velocity, like speed, has SI units of meters per second (m/s), but because it is a vector, you must also include a direction. Furthermore, the variable **v** for velocity is bold because it is a vector, which is in contrast to the variable *v* for speed which is italicized because it is a scalar quantity.

### Tips For Success

It is important to keep in mind that the average speed is not the same thing as the average velocity without its direction. Like we saw with displacement and distance in the last section, changes in direction over a time interval have a bigger effect on speed and velocity.

Suppose a passenger moved toward the back of a plane with an average velocity of –4 m/s. We cannot tell from the average velocity whether the passenger stopped momentarily or backed up before he got to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals such as those shown in Figure 2.9. If you consider infinitesimally small intervals, you can define instantaneous velocity, which is the velocity at a specific instant in time. Instantaneous velocity and average velocity are the same if the velocity is constant.

Earlier, you have read that distance traveled can be different than the magnitude of displacement. In the same way, speed can be different than the magnitude of velocity. For example, you drive to a store and return home in half an hour. If your car’s odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero because your displacement for the round trip is zero.

### Watch Physics

#### Calculating Average Velocity or Speed

This video reviews vectors and scalars and describes how to calculate average velocity and average speed when you know displacement and change in time. The video also reviews how to convert km/h to m/s.

Grasp Check

Which of the following fully describes a vector and a scalar quantity and correctly provides an example of each?

- A scalar quantity is fully described by its magnitude, while a vector needs both magnitude and direction to fully describe it. Displacement is an example of a scalar quantity and time is an example of a vector quantity.
- A scalar quantity is fully described by its magnitude, while a vector needs both magnitude and direction to fully describe it. Time is an example of a scalar quantity and displacement is an example of a vector quantity.
- A scalar quantity is fully described by its magnitude and direction, while a vector needs only magnitude to fully describe it. Displacement is an example of a scalar quantity and time is an example of a vector quantity.
- A scalar quantity is fully described by its magnitude and direction, while a vector needs only magnitude to fully describe it. Time is an example of a scalar quantity and displacement is an example of a vector quantity.

### Worked Example

#### Calculating Average Velocity

A student has a displacement of 304 m north in 180 s. What was the student's average velocity?

### Strategy

We know that the displacement is 304 m north and the time is 180 s. We can use the formula for average velocity to solve the problem.

Solution

2.1$${v}_{\text{avg}}=\frac{\Delta d}{\Delta t}=\frac{304\text{m}}{180\text{s}}=1.7\text{m/snorth}$$

Discussion

Since average velocity is a vector quantity, you must include direction as well as magnitude in the answer. Notice, however, that the direction can be omitted until the end to avoid cluttering the problem. Pay attention to the significant figures in the problem. The distance 304 m has three significant figures, but the time interval 180 s has only two, so the quotient should have only two significant figures.

### Tips For Success

Note the way scalars and vectors are represented. In this book d represents distance and displacement. Similarly, v represents speed, and v represents velocity. A variable that is not bold indicates a scalar quantity, and a bold variable indicates a vector quantity. Vectors are sometimes represented by small arrows above the variable.

### Worked Example

#### Solving for Displacement when Average Velocity and Time are Known

Layla jogs with an average velocity of 2.4 m/s east. What is her displacement after 46 seconds?

### Strategy

We know that Layla's average velocity is 2.4 m/s east, and the time interval is 46 seconds. We can rearrange the average velocity formula to solve for the displacement.

Solution

2.2$$\begin{array}{ccc}{v}_{\text{avg}}& =& \frac{\Delta d}{\Delta t}\hfill \\ \hfill \Delta d& =& {v}_{avg}\Delta t\hfill \\ & =& (2.4\text{m/s)(46s)}\hfill \\ & =& 1.1\times {10}^{2}\text{meast}\hfill \end{array}$$

Discussion

The answer is about 110 m east, which is a reasonable displacement for slightly less than a minute of jogging. A calculator shows the answer as 110.4 m. We chose to write the answer using scientific notation because we wanted to make it clear that we only used two significant figures.

### Tips For Success

Dimensional analysis is a good way to determine whether you solved a problem correctly. Write the calculation using only units to be sure they match on opposite sides of the equal mark. In the worked example, you have

m = (m/s)(s). Since seconds is in the denominator for the average velocity and in the numerator for the time, the unit cancels out leaving only m and, of course, m = m.

### Worked Example

#### Solving for Time when Displacement and Average Velocity are Known

Phillip walks along a straight path from his house to his school. How long will it take him to get to school if he walks 428 m west with an average velocity of 1.7 m/s west?

### Strategy

We know that Phillip's displacement is 428 m west, and his average velocity is 1.7 m/s west. We can calculate the time required for the trip by rearranging the average velocity equation.

Solution

2.3$$\begin{array}{ccc}\hfill {v}_{\text{avg}}& =& \frac{\Delta d}{\Delta t}\hfill \\ \hfill \Delta t& =& \frac{\Delta d}{{v}_{\text{avg}}}\hfill \\ & =& \frac{428\text{m}}{1.7\text{m/s}}\hfill \\ & =& 2.5\times {10}^{2}\text{s}\hfill \end{array}$$

Discussion

Here again we had to use scientific notation because the answer could only have two significant figures. Since time is a scalar, the answer includes only a magnitude and not a direction.