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We have noted that distance traveled can be greater than displacement. So average speed can be greater than average velocity, which is displacement divided by time. 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 12 km/h. 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.

A house and a store, with a set of arrows in between showing that the distance between them is 3 point 0 kilometers and the total distance traveled, delta x total, equals 0 kilometers.
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.

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 [link] . (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.)

Three line graphs. First line graph is of position in kilometers versus time in hours. The line increases linearly from 0 kilometers to 6 kilometers in the first 0 point 25 hours. It then decreases linearly from 6 kilometers to 0 kilometers between 0 point 25 and 0 point 5 hours. Second line graph shows velocity in kilometers per hour versus time in hours. The line is flat at 12 kilometers per hour from time 0 to time 0 point 25. It is vertical at time 0 point 25, dropping from 12 kilometers per hour to negative 12 kilometers per hour. It is flat again at negative 12 kilometers per hour from 0 point 25 hours to 0 point 5 hours. Third line graph shows speed in kilometers per hour versus time in hours. The line is flat at 12 kilometers per hour from time equals 0 to time equals 0 point 5 hours.
Position vs. time, velocity vs. time, and speed vs. time on a trip. Note that the velocity for the return trip is negative.

A commuter train travels from Baltimore to Washington, DC, and back in 1 hour and 45 minutes. The distance between the two stations is approximately 40 miles. What is (a) the average velocity of the train, and (b) the average speed of the train in m/s?

(a) The average velocity of the train is zero because x f = x 0 size 12{x rSub { size 8{f} } =x rSub { size 8{0} } } {} ; the train ends up at the same place it starts.

(b) The average speed of the train is calculated below. Note that the train travels 40 miles one way and 40 miles back, for a total distance of 80 miles.

distance time = 80 miles 105 minutes size 12{ { {"distance"} over {"time"} } = { {"80 miles"} over {"105 minutes"} } } {}
80 miles 105 minutes × 5280 feet 1 mile × 1 meter 3 . 28 feet × 1 minute 60 seconds = 20 m/s

Section summary

  • Time is measured in terms of change, and its SI unit is the second (s). Elapsed time for an event is
    Δ t = t f t 0 , size 12{Δt=t rSub { size 8{f} } - t rSub { size 8{0} } } {}
    where t f size 12{t rSub { size 8{f} } } {} is the final time and t 0 size 12{t rSub { size 8{0} } } {} is the initial time. The initial time is often taken to be zero, as if measured with a stopwatch; the elapsed time is then just t size 12{t} {} .
  • Average velocity v - size 12{ { bar {v}}} {} is defined as displacement divided by the travel time. In symbols, average velocity is
    v - = Δ x Δ t = x f x 0 t f t 0 . size 12{ { bar {v}}= { {Δx} over {Δt} } = { {x rSub { size 8{f} } - x rSub { size 8{0} } } over {t rSub { size 8{f} } - t rSub { size 8{0} } } } "." } {}
  • The SI unit for velocity is m/s.
  • Velocity is a vector and thus has a direction.
  • Instantaneous velocity v size 12{v} {} is the velocity at a specific instant or the average velocity for an infinitesimal interval.
  • Instantaneous speed is the magnitude of the instantaneous velocity.
  • Instantaneous speed is a scalar quantity, as it has no direction specified.
  • Average speed is the total distance traveled divided by the elapsed time. (Average speed is not the magnitude of the average velocity.) Speed is a scalar quantity; it has no direction associated with it.

Conceptual questions

There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.

Does a car’s odometer measure position or displacement? Does its speedometer measure speed or velocity?

If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you calculating the average speed or the magnitude of the average velocity? Under what circumstances are these two quantities the same?

How are instantaneous velocity and instantaneous speed related to one another? How do they differ?

Problems&Exercises

(a) Calculate Earth’s average speed relative to the Sun. (b) What is its average velocity over a period of one year?

(a) 3 . 0 × 10 4 m/s size 12{3 "." "0 " times "10" rSup { size 8{4} } " m/s"} {}

(b) 0 m/s

A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the average speed of the blade tip in the helicopter’s frame of reference. (b) What is its average velocity over one revolution?

The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present?

2 × 10 7 years size 12{2 times "10" rSup { size 8{7} } " years"} {}

The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light ( 3 . 00 × 10 8 m/s ) size 12{ \( 3 "." "00" times "10" rSup { size 8{8} } " m/s" \) } {} .

384,000 km

A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity (a) for each of the three intervals and (b) for the entire motion.

Practice Key Terms 7

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Source:  OpenStax, Kinematics. OpenStax CNX. Sep 11, 2015 Download for free at https://legacy.cnx.org/content/col11878/1.5
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