Velocity or speed? Instantaneous or average? Keep building your physics vocabulary.

What does velocity mean?

Your notion of velocity is probably similar to its scientific definition. You know that a large displacement in a small amount of time means a large velocity and that velocity has units of distance divided by time, such as miles per hour or kilometers per hour.
Average velocity is defined to be the change in position divided by the time of travel.
vavg=ΔxΔt=xfx0tft0\Large v_{avg}=\dfrac{\Delta x}{\Delta t}=\dfrac{x_f-x_0}{t_f-t_0}
In this formula, vavgv_{avg} is the average velocity; Δx\Delta x is the change in position, or displacement; and xfx_f and x0x_0 are the final and beginning positions at times tft_f and t0t_0, respectively. If the starting time t0t_0 is taken to be zero, then the average velocity is written as below:
vavg=Δxtv_{avg}=\dfrac{\Delta x}{t}
Note: tt is shorthand for Δt\Delta t.
Notice that this definition indicates that velocity is a vector because displacement is a vector. It has both magnitude and direction. The International System of Units (SI) unit for velocity is meters per second or ms\dfrac{\text{m}}{\text{s}}, but many other units such as kmhr\dfrac{\text{km}}{\text{hr}}, mihr\dfrac{\text{mi}}{\text{hr}} (also written as mph), and cms\dfrac{\text{cm}}{\text{s}} are commonly used. Suppose, for example, an airplane passenger took 5 seconds to move −4 meters, where the negative sign indicates that displacement is toward the back of the plane. His average velocity can be written as below:
vavg=Δxt=4 m5 s=0.8msv_{avg}=\dfrac{\Delta x}{t}=\dfrac{-4\text{ m}}{5 \text{ s}}=-0.8 \dfrac{\text m}{\text{s}}
The minus sign indicates the average velocity is also toward the rear of the plane.
The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point, however. For example, we cannot tell from average velocity whether the airplane passenger stops momentarily or backs up before he goes to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals. For instance, in the figure below, we see that the total trip displacement, Δxtot\Delta x _ \text{tot}, consists of 4 segments, Δxa\Delta x_\text a, Δxb\Delta x_\text b, Δxc\Delta x_\text c, and Δxd\Delta x_\text d.
Figure 1: A more detailed record of an airplane passenger heading toward the back of the plane, showing smaller segments of his trip. (Image credit: Openstax College Physics)
The smaller the time intervals considered in a motion, the more detailed the information. Carrying this process to its logical conclusion, we are left with an infinitesimally small interval. Over such an interval, the average velocity becomes the instantaneous velocity, or the velocity at a specific moment. A car’s speedometer, for example, shows the magnitude—but not the direction—of the instantaneous velocity of the car. Police give tickets based on instantaneous velocity, but when calculating how long it will take to get from one place to another on a road trip, you need to use average velocity. Instantaneous velocity, vv, is simply the average velocity at a specific instant in time or over an infinitesimally small time interval.
Mathematically, finding instantaneous velocity, vv, at a precise instant tt can involve taking a limit, a calculus operation beyond the scope of this article. However, under many circumstances, we can find precise values for instantaneous velocity without calculus.

What does speed mean?

In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not have the same meaning, and they are distinct concepts. One major difference is that speed has no direction. Thus, speed is a scalar. Just as we need to distinguish between instantaneous velocity and average velocity, we also need to distinguish between instantaneous speed and average speed.
Instantaneous speed is the magnitude of instantaneous velocity. For example, suppose the airplane passenger at one instant had an instantaneous velocity of , the negative meaning toward the rear of the plane. At that same time his instantaneous speed was 3.0ms3.0 \dfrac{\text {m}}{\text{s}}. Or suppose that at a particular instant during a shopping trip, your instantaneous velocity is 40kmhr40 \dfrac{\text {km}}{\text{hr}} due north. Your instantaneous speed at that instant would be 40kmhr40 \dfrac{\text {km}}{\text{hr}}—the same magnitude but without a direction. Average speed, however, is very different from average velocity. Average speed is the distance traveled divided by elapsed time. So, while the magnitudes of the instantaneous speed and velocity are always identical, the magnitudes of average speed and velocity can be very different.
Since distance traveled can be greater than the magnitude of displacement, the average speed can be greater than the magnitude of the average velocity. For example, if you drive to a store and return home in half an hour and your car’s odometer shows the total distance traveled was 6 km, then your average speed was 12kmhr12 \dfrac{\text {km}}{\text{hr}}. Your average velocity, however, was zero because your displacement for the round trip is zero—Displacement is change in position and, thus, is zero for a round trip. Thus average speed is not simply the magnitude of average velocity.
Figure 2: During a 30-minute round trip to the store, the total distance traveled is 6 km. The average speed is 12 km/h. The displacement for the round trip is zero, since there was no net change in position. Thus the average velocity is zero. Image credit: Openstax College Physics
Another way of visualizing the motion of an object is to use a graph. A plot of position or of velocity as a function of time can be very useful. For example, for this trip to the store, the position, velocity, and speed-vs.-time graphs are displayed in Figure 3. Note that these graphs depict a very simplified model of the trip. We are assuming that speed is constant during the trip, which is unrealistic given that we’ll probably stop at the store. But for simplicity’s sake, we will model it with no stops or changes in speed. We are also assuming that the route between the store and the house is a perfectly straight line.
Figure 3: Position vs. time, velocity vs. time, and speed vs. time on a trip. Note that the velocity for the return trip is negative. Image credit: Openstax College Physics)

What do solved examples involving velocity and speed look like?

Example 1: Disoriented iguana

An iguana with a poor sense of spatial awareness is walking back and forth in the desert. First the iguana walks 12 meters to the right in a time of 20 seconds. Then the iguana runs 16 meters to the left in a time of 8 seconds.
What was the average speed and average velocity of the iguana for the entire trip?
Assume that rightward is the positive direction.
To find the average speed we take the total distance traveled divided by the time interval.
average speed=distance traveledtime interval=12.0 m+16.0 m20.0 s+8.0 s\text{average speed}=\dfrac{\text{distance traveled}}{\text{time interval}}=\dfrac{12.0\text{ m}+16.0\text{ m}}{20.0\text{ s}+8.0\text{ s}}
average speed=28.0 m28.0 s\text{average speed}=\dfrac{28.0\text{ m}}{28.0\text{ s}}
average speed=1 m s\text{average speed}=1\dfrac{\text{ m}}{\text{ s}}
To find the average velocity we take the displacement Δx\Delta x divided by the time interval.
average velocity=displacementtime interval=4.0 m28.0 s\text{average velocity}=\dfrac{\text{displacement}}{\text{time interval}}=\dfrac{-4.0\text{ m}}{28.0\text{ s}}
average velocity=17 m s\text{average velocity}=-\dfrac{1}{7}\dfrac{\text{ m}}{\text{ s}}
The iguana walked 12 meters to the right and then ran 16 meters to the left. So the iguana ended up 4 meters to the left of its starting point.
This means the magnitude of the displacement for the entire trip was 4 meters.

Example 2: Hungry dolphin

A hungry dolphin is swimming horizontally back and forth looking for food. The motion of the dolphin is given by the position graph shown below.
Determine the following for the dolphin:
a. average velocity between time t=0 st=0 \text{ s} to t=6 st=6\text{ s}
b. average speed between t=0 st=0 \text{ s} to t=6 st=6\text{ s}
c. instantaneous velocity at time t=1 st=1\text{ s}
d. instantaneous speed at time t=4 st=4\text{ s}
Part A: Average velocity is defined to be the displacement per time.
vavg=ΔxΔt=0 m8 m6 s0 s=8 m6 s(Use definition of average velocity.)v_{avg}=\dfrac{\Delta x}{\Delta t}=\dfrac{0\text { m}-8\text{ m}}{6 \text{ s}-0\text{ s}}=\dfrac{-8\text{ m}}{6 \text{ s}}\quad \text{(Use definition of average velocity.)}
vavg=43ms(Calculate and celebrate.)v_{avg}=-\dfrac{4}{3} \dfrac{\text m}{\text s}\quad \text{(Calculate and celebrate.)}
Part B: Average speed is defined to be the distance traveled per time. The distance is the sum of the total path length traveled by the dolphin, so we just add up all the distances traveled by the dolphin for each leg of the trip.
vavg=distance traveledΔt=12 m+0 m+4 m6 s0 s=16 m6 s(use definition of average speed)v_{avg}=\dfrac{\text{distance traveled}}{\Delta t}=\dfrac{12\text{ m}+0\text{ m}+4\text{ m}}{6 \text{ s}-0\text{ s}}=\dfrac{16\text{ m}}{6 \text{ s}}\quad\text{(use definition of average speed)}
Between t=0st=0\text{s} and t=3st=3\text{s}, the dolphin swam a distance of 12 meters since it went from x=8mx=8\text{m} to x=4mx=-4\text{m}.
Then from t=3st=3\text{s} to t=5st=5\text{s}, the dolphin swam a distance of 0 meters since it remained at x=4mx=-4\text{m} and didn't move.
Then from t=5st=5\text{s} to t=6st=6\text{s}, the dolphin swam a distance of 4 meters since it went from x=4mx=-4\text{m} to x=0mx=0\text{m}.
So the total distance traveled by the dolphin was 12 m+0 m+4 m=16 m12\text{ m}+0\text{ m}+4\text{ m}=16\text{ m}.
vavg=83ms(calculate and celebrate)v_{avg}=\dfrac{8}{3} \dfrac{\text m}{\text s}\quad \text{(calculate and celebrate)}
Part C: Instantaneous velocity is the velocity at a given moment and will be equal to the slope of the graph at that moment. To find the slope at t=1st=1\text{s} we can determine the "rise over run" for any two points on the graph between t=0st=0\text{s} and t=3st=3\text{s} (since the slope is constant between those times). Choosing the times t=2st=2\text{s} and t=0st=0\text{s}, we find the slope as follows,
vinstantaneous=0 m8 m2 s0 s=8 m2 sv_\text{instantaneous}=\dfrac{0\text{ m}-8\text{ m}}{2\text{ s}-0\text{ s}}=\dfrac{-8\text{ m}}{2\text{ s}}
The position at t=2st=2\text{s} is x2=0mx_2=0\text{m}, and the position at t=0 st=0\text{ s} is x0=8 mx_0=8\text{ m}.
So between those two times
Δx=x2x0\Delta x=x_2-x_0
Δx=0 m8 m\Delta x=0\text{ m}-8\text{ m}
Δx=8 m\Delta x=-8\text{ m}
The time interval was 2 seconds, so the slope and velocity were
v=8 m2 sv=\dfrac{-8\text{ m}}{2\text{ s}}
Part D: Instantaneous speed is the speed at a given moment in time and will be equal to the magnitude of the slope. Since the slope at t=4st=4\text{s} is equal to zero, the instantaneous speed at t=4st=4\text{s} is also equal to zero.
This article was adapted from the following article:
  1. "Time, Velocity, and Speed" from Openstax College Physics. Download the original article free at