Average speed, average velocity, and instantaneous velocity

Transcription

1 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

2 2 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 (0.18 m/s). Overall, the average speed for the trip was total distance traveled total time required for the travel (E3.1.1) So, in this case, 3.0 meters meters meters ( ) seconds = 1 5 meter per second = 0.20 m/s. = 12 meters 60 seconds Note that the average speed for the entire trip is not the average of the average speeds during each of the three parts, 1 3 ( ) meters per second (which is 0.21 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,

3 3 then halted for the last 20 minutes as a patrolman gave the driver a ticket (picture above). Its average speed really does not describe that hour s journey very well. The way to deal with such a situation is to take the average over a shorter interval, and do this more times. In the example given above, if the speed were sampled every 15 minutes instead of over the entire hour, we would list average speeds of 150 km/h at the start (0 to 15 minutes), 150 km/h (15 to 30 minutes), 100 km/h (30 to 45 minutes), and 0 km/h (45 to 60 minutes). Why is the third entry 100 km/h? During the 30 to 45 minute interval, the car traveled at 150 km/h for only 10 minutes ( 1 6 of an hour), between 30 min and 40 min, traveling during that time a distance of distance = speed x time = 150 km/h x 1 6 h = km = 25 km. During the remainder of this time interval it did not move, so it traveled 0 km. The total distance traveled during this 15 minute interval is then 25 km + 0 km = 25 km. However, the average is taken over the entire 15 minutes, or 1 4 of an hour, so the average speed (Eq. E3.1.1) is

4 4 distance traveled time required = 25 km 1 = 100 km/h. 4 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

5 5 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 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, km 6.52 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 880 km 9 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

6 6 restrooms. In this case, the average speed while driving would again be found by using the definition, taking the total time while driving as the denominator rather than the total time. We find v av while driving = 880 km 8.67 h = km/h. This is just below the 65 mi/h speed limit (104 km/h). Calculating average velocity from net displacement and time The average velocity is a vector quantity, which means that it has a direction as well as a numerical value. A vector quantity is shown by an arrow over the symbol ( v ) or by making the symbol boldface (v). We define average velocity, v av, as the net displacement (displacement is the change in position, which includes both distance and direction) divided by the total time required to achieve that displacement: displacement total time required for the travel (E3.1.2) For example, a car traveling 100 kilometers northward in an hour will have an average velocity of 100 km/h, north. A car traveling 55 miles northwest in an hour would have an average velocity of 55 mi/h, northwest. Consider again the example of the pig waddling 3.0 meters eastward in 12 seconds, 4.0 meters northward in 20 seconds, and 5.0 meters at 53.1 south of west back to the original spot in the mud puddle in 28 seconds (see the picture on the next page).

7 7 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 east = 0.25 m/s, east; 12 s 4.0 m north v av, second segment = = 0.20 m/s, north; and 20 s 5.0 m at 53.1 south of west v av, third segment = = 0.18 m/s, 53.1 south of west. 28 s 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

8 8 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) 940 million km 31.6 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 Thus distance traveled = z final - z 0 = z = 12.0 m, up m, up = 6.0 m, up. z = m or 6.0 m, up. Consequently, from the definition E3.1.2, 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.

9 9 Since the velocity is constant, the average velocity is the constant velocity, 3.00 m/s, east. The average velocity is the change in displacement, r, divided by the time required to make that displacement, t, so we can find the change in displacement as r = v t = [3.00 m/s, east][30.0 s] = 90.0 m, east. This change in displacement must be added to the original position to produce the new position. Thus, the new position is 5.0 m, west m, east = 85.0 m, east. Instantaneous velocity and how to calculate it We may define the instantaneous velocity of an object to be the velocity at a particular instant of time. In reality, we must find this instantaneous velocity by measuring successive positions. Then we must keep calculating average velocities over smaller and smaller time intervals. Look at Fig. E3.1.1 on the following page, which shows the position of a moving object between times 0 and 20 seconds. In order to make the graph, we had to set up a coordinate system along which to measure positions. This picks out a specific direction in space. Suppose we wanted to know the instantaneous speed at the time 17 s. The positions at every second have been highlighted in Fig. E3.1.1, so we know the positions at all times, but not the velocities. Because the position is given, the change in position can be found in magnitude and in direction; this allows velocity to be calculated. If we ignored the position information, we d be calculating the speed instead of the velocity of the object.

10 Position (m) Time (s) 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 m = + 62 m = m/s. 20 s - 14 s 6 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 m = + 41 m = m/s. 19 s - 15 s 4 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

11 11 v av 101 m - 80 m = + 21 m = m/s. 18 s - 16 s 2 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 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 32 m - 0 m av 10 s - 0 s = +32 m = +3.2 m/s, 10 s from 1 to 9 s, positions around +1 m and +26 m, for an average velocity v 26 m - 1 m av 9 s - 1 s = + 25 m 8 s = m/s, from 2 to 8 s, positions around +2 m and +21 m, for an average velocity of about v 21 m - 2 m av 8 s - 2 s = + 19 m 6 s = m/s, from 3 to 7 s, positions around +3 m and +16 m, for an average velocity of about v 16 m - 3 m av 7 s - 3 s = + 13 m 4 s = m/s, from 4 to 6 s, positions around +6 m and +12 m, for an average velocity of about v 12 m - 6 m av 6 s - 4 s = + 6 m 2 s = +3.0 m/s, and for smaller and smaller intervals finally becoming the instantaneous velocity of +3.2 m/s.

12 12 The numbers were closer for the earlier time because the changes in the position were not so great as occurred at the later time. We have chosen to take the interval equally spaced around the target time at which we want the velocity, but we need not have done so. Suppose we d chosen to go from 17 s to 20 s (then 19, then 18) to get the average velocity at 17 s. You can verify that the respective average velocities would have been +124 m - 90 m 20 s - 17 s = m/s, 112 m - 90 m = m/s, and 19 s - 17 s 101 m - 90 m = m/s. 18 s - 17 s We need to get closer, much closer, to the 17 s mark to find the instantaneous speed of m/s, but it still eventually happens.