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Average speed, average velocity, and instantaneous velocity
Calculating average speed from distance traveled and time
We define the average speed, v av, as the total distance traveled divided by the time required to travel that distance. The unit of speed is the meter per second (m/s) or the kilometer per hour (km/h) [and, in America’s customary system of measurement, the foot per second (ft/s) or the mile per hour (mi/h)]. If an object travels 50 meters in 10 seconds, it has an average speed of 50 meters/10 seconds, or 5 meters per second. If an object travels a distance of 100 kilometers in the course of an hour, it has an average speed of 100 kilometers per hour. If an object travels 45 miles in the course of an hour, it has an average speed of 45 miles per hour.
Suppose a pig arises from a mud puddle and waddles 3.0 meters eastward in a straight line, then 4.0 meters northward in a straight line, then 5.0 meters in a straight line back to the original spot from which he started (see the picture above). Suppose also that the 3- meter walk took 12 seconds, the 4-meter walk 20 seconds, and the 5-meter walk 28 seconds. During the first part of the walk, the average speed is 3.0 meters/12 seconds, or
1/4 meters per second (0.25 m/s). During the second part of the walk, the average speed is 4.0 meters/20 seconds, or 1/5 meters per second (0.20 m/s). On the pig’s return to the mud puddle, the average speed is 5.0 meters/28 seconds, or 5/28 meters per second (0. m/s). Overall, the average speed for the trip was
v av = (^) total time required for the traveltotal distance traveled (E3.1.1)
So, in this case, v av = 3.0 meters + 4.0 meters + 5.0 meters(12 + 20 + 28) seconds = 60 seconds12 meters = 15 meter per second = 0.20 m/s.
Note that the average speed for the entire trip is not the average of the average speeds during each of the three parts, 13 (0.25 + 0.20 + 0.18) meters per second (which is 0. m/s), because each of the time intervals is different.
Average speed is useful in describing the motion of a body, but it does not tell everything about the motion. For instance, a car traveling at an average speed of 100 kilometers per hour during a certain hour could conceivably have traveled at 150 km/h for 40 minutes,
v av = distance traveledtime required = 25 km 1 4 h^
= 100 km/h.
If the speed were sampled every 5 minutes, we would list average speeds of 150 km/h at the start, 150 km/h at 0 to 5, 150 km/h at 5 to 10, 150 km/h at 10 to 15, 150 km/h at 15 to 20, 150 km/h at 20 to 25, 150 km/h at 25 to 30, 150 km/h at 30 to 35, 150 km/h at 35 to 40, 0 km/h at 40 to 45, 0 km/h at 45 to 50, 0 km/h at 50 to 55, and 0 km/h at 55 to 60 minutes. Even though such a list is tedious to read, it certainly represents the car’s journey more accurately than an hour’s or fifteen minutes’ average. This is about the same as the first picture.
For an object’s motions that are more variable than in the idealizations so far presented, we are forced to take samples over smaller and smaller time intervals. The more varied the motion is, the more a greater number of samples is needed to get an accurate record of that motion. Of course, no matter how small the time interval, we can find the speed if we can determine the distance traveled. The instantaneous speed is the speed of any object at an instant, which we visualize as being the distance traveled in a very short time divided by
that time interval. As we make the time interval shorter and shorter, we approach nearer and nearer to the true speed in an interval, the instantaneous speed.
Let’s take another example and calculate the speed. Your car travels 517.7 km in 6.52 h of driving. What was your average speed? The average speed is the distance traveled divided by the time interval during which the travel was accomplished. Therefore, from Eq. E3.1.1, v av = 517.7 km6.52 h = 79.4 km/h.
Could you have been stopped for speeding during that time if the car was driving in a 55 mi/h zone? This corresponds to a speed limit of about 90 km/h. The 79.4 km/h is below the speed limit, but since all we can calculate is the average speed, the car could have exceeded the speed limit. For example, if you stopped for an hour to eat during the drive, your average speed while you were driving could have been substantially higher, about 100 km/h. Even if you didn’t stop, you could have exceeded the speed limit many times. This speeding question cannot really be answered definitively without more information.
Sometimes my family drives from our home in Delaware, Ohio to visit my brother in Montclair, New Jersey. Usually, we try to drive without stopping except once for gas. The distance is just about 880 km/h and it takes us a shade under 9 hours to drive there. What is the average speed for the trip? What is the average speed while we are driving?
Again using the definition E3.1.1, the average speed for the trip is
v av = 880 km9 h = 97.8 km/h.
We must assume something about our stop for gas in order to find the average speed while we were driving. It’s probable that it took 20 minutes to fill the tank and visit the
Since the motion in each part of the pig’s walk was in a straight line, the velocities for each part have lengths given by the respective average speeds. Thus, the three average velocities are, given the definition of Eq. E3.1.2:
→ v av, first segment = 3 m east12 s = 0.25 m/s, east; → v av, second segment = 4.0 m north20 s = 0.20 m/s, north; and → v av, third segment = 5.0 m at 53.1° south of west28 s = 0.18 m/s, 53.1° south of west.
The average velocity for the trip is, however, given by the net displacement, zero, divided by 60 seconds, because the motion is not just in a straight line. The net displacement of the ending point with respect to the starting point is zero, even though the pig travels a path that is a total of 12 meters long. As a result, the average velocity is zero (while the average speed is 0.21 m/s, as found already).
Anything that returns to its starting position after some specified time will have an average velocity of zero over that time interval, because its net displacement is zero, no matter how great a distance it traveled in the meantime. Earth travels in an almost circular orbit around the Sun and takes exactly 1 year to return to its starting position relative to
the Sun. Thus Earth’s average velocity with respect to the Sun, measured over 1 year, is zero, despite the fact that Earth has traveled a distance of about 940 million kilometers during that time. Its average speed is not zero; it is (using Eq. E3.1.1)
v av = 940 million km31.6 million s = 29.7 km/s.
An otherwise motionless person is standing in an elevator that is rising. At the time we set to be zero, the person is standing in the elevator at rest on the third floor, 6.0 m up from the ground. After another 10.0 s, the person is standing in the elevator at rest on the fifth floor, 12.0 m up. How do we find the person’s average velocity over the 10.0 second interval?
While the motion takes place in three dimensions, the person is not moving except for the elevator’s motion, the displacement is zero for every direction but up. The displacement in the up-direction (call this direction z) is distance traveled = zfinal - z 0 = ∆z = 12.0 m, up - 6.0 m, up = 6.0 m, up.
Thus ∆z = + 6.0 m or 6.0 m, up. Consequently, from the definition E3.1.2,
→ v av = 0.60 m/s, up.
Let’s use the velocity to find the displacement. Suppose that an object has a constant velocity of 3.00 m/s, east (as measured in some system of coordinates). What is its displacement from the origin of the coordinate system—its position—after a time 30.0 s? The original position is 5.0 m, west.
Time (s)
Position (m)
0
20
40
60
80
100
120
140
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Fig. E3.1.1 The positions of an object at one-second intervals between times 0 and 20 s.
Using the definition of average velocity, we might average over the time interval from 14 s to 20 s, which, since the positions at these times are about +62 m and +124 m, respectively, gives
→ v av^ 124 m - 62 m20 s - 14 s = +62 m6 s = +10.33 m/s.
The plus sign indicates the direction, the direction defined as positive. If we shrank the range and used 15 s to 19 s, we’d get instead, positions of about +71 m and +112 m, and so → v av^ 112 m - 71 m19 s - 15 s = +41 m4 s = +10.25 m/s.
If we’d used the even smaller interval from 16 s to 18 s, positions of about +80 m and +101 m, we’d have gotten
→ v av^ 101 m - 80 m18 s - 16 s = +21 m2 s = +10.50 m/s.
Obviously, your eyes might indicate slightly different values for the positions at each of the times from your reading of the graph (but not very much different). It’s even harder to read the values from the graph for shorter intervals than 1 s, but if we were able to discern them, we’d get closer and closer to +10.40 m/s the shorter the interval we chose.
With this graph, note that the average speed and average velocity are different around different times. Had we chosen to find the average velocity at 5 s instead, we’d have found from 0 to 10 s positions around 0 m and 32 m, or
→ v av^ 32 m - 0 m10 s - 0 s = +32 m10 s = +3.2 m/s,
from 1 to 9 s, positions around +1 m and +26 m, for an average velocity
→ v av^ 26 m - 1 m9 s - 1 s = +25 m8 s = +3.25 m/s,
from 2 to 8 s, positions around +2 m and +21 m, for an average velocity of about
→ v av^ 21 m - 2 m8 s - 2 s = +19 m6 s = +3.33 m/s,
from 3 to 7 s, positions around +3 m and +16 m, for an average velocity of about
→ v av^ 16 m - 3 m7 s - 3 s = +13 m4 s = +3.25 m/s,
from 4 to 6 s, positions around +6 m and +12 m, for an average velocity of about
→ v av^ 12 m - 6 m6 s - 4 s = +6 m2 s = +3.0 m/s,
and for smaller and smaller intervals finally becoming the instantaneous velocity of +3. m/s.
