Speed and Velocity

Have you noticed that scalar quantities ignore direction, and vector quantities take direction into account? Speed and velocity also sound the same, don’t they? But again, one is a vector and one is a scalar. Speed is the scalar quantity that describes how fast something is moving, like 100 mph. It’s the rate that something covers over a distance.

Rockets are fast, so they have high speeds, which means they cover large distances in a short amount of time. Compared to the speed of light, however, rockets are quite slow. (You always have to keep in mind what you are comparing to.) Velocity is a vector quantity that has a magnitude and a direction, like 100 mph north. It doesn’t matter if your speeding up or slowing down (we take that into account when we look at acceleration of an object). Velocity is the change in distance over a given time, or v = d / t. If a jet travels 600 miles in an hour, then it’s moving at 600 mph. A car going 25 miles in a half hour is moving at 50 mph. A snail crawling an inch every four minutes is moving at 0.25 inches per minute. You can mix up the units of distance and time to be whatever is most useful to you, whether it’s miles per hour, feet per minute, or meters per second. Most objects don’t just travel at one speed, however.

When you travel in a car, sometimes it’s on the freeway (65 mph), sometimes you’re at a stoplight (zero mph), sometimes you’re driving through the neighborhood (25 mph), and so forth. Your car has a lot of speed changes, so it’s useful to be able to calculate the average speed and average velocity of your car. It’s also useful to know the speed or velocity at a given instant in time, called your instantaneous speed or instantaneous velocity.

Please login or register to read the rest of this content.


13 Responses to “Speed and Velocity”
  1. Thanks for the update!

  2. Ramalogos says:

    Every thing is working fine now. Don’t know what happened.

  3. Ramalogos says:

    A little problem…When I try to go back a little to hear you or see what you wrote, it jumps way forward, then I have to start all over again just to get to the part I need to hear/see again.

  4. Aurora says:

    The exercises are the last link in the navigation on the right hand side of the page. Here’s a direct link:


  5. laurene_ross says:

    I was wondering if there was homework pages for this lesson, if there are any where would they be located.

  6. Aurora says:

    Velocity is speed and direction, like 65 mph north. Rate… of what? You could saw speed is how fast (or slow) you cover a distance is per unit of time, so its the change of rate of distance per time. When you change the velocity, now you have acceleration. There’s a whole unit on this right where you are at, because it’s a bit confusing!

  7. Karen Daley says:

    Is velocity the same thing as rate?

  8. Aurora says:

    Maybe I am not understanding your question right (which can easily happen when we’re not face-to-face and I am sitting right next to you, looking at your work). Let’s try again? 🙂

    Average speed is the distance traveled over the time elapsed. It takes me four hours to drive 200 miles, so I have an average speed of 50 miles per hour. It’s a scalar quantity.

    Since velocity is a vector, the average velocity is the displacement over time for two specific points like this:
    Vavg= (x2 – x1) / (t2 – t1) for a straight line.

    If acceleration is constant (you are not speeding up or slowing down), and you know the velocity when you started and ended, then the average velocity is the sum of the velocities divided by 2. (This is a special case, and you can see this will give you the same result as the equation above.)

    If you are changing speeds (non-constant acceleration), then the average (vector) velocity quantity then you must use the distance (as a vector) divided by time. Simply averaging the velocity vectors will not work, since the velocities are in different directions and the acceleration is not constant.

  9. Rosalind Hitchcock says:

    Thanks Aurora. Still struggling a bit with this … For the example I gave in my previous post (travel of 60mph for 1 hour, then 30mph for 1 second), the average speed would be 45mph? IOW, time is completely ignored when calculating average speed – is that right? That seems inconsistent with the formulas given for average speed, all of which seem to include time (e.g., xf – xi = 1/2 (vf + vi) t, and also seems really counterintuitive, but if we need to re-train our intuition, I’m up for that 🙂

    We’re just getting started, and already stuck! 🙁 Thanks again for your help.

  10. Aurora says:

    That’s true – you do need more data points to get a more accurate average. However if you only have 2 points, then that’s what you use. (you need at least 30 data points to be statically accurate for a set of conditions, or three per trial to be scientifically accurate.)

    For the worrier problem:

    V1 = -10 ft/min
    V2 = 12.5 ft/min
    V3 = -10 ft/min

    Average VELOCITY is -7.5 ft / min (you must account for direction)
    Average SPEED is (10 + 12.5 + 10) / 3 = 11 ft / min

    You can’t separate distance and time when doing your calculation – it’s not accurate since it doesn’t account for the distance traveled in which unit of time. You’re taking the averages of three numbers, not six, to get the average speed.

  11. Rosalind Hitchcock says:

    A question re: the average velocity problem with the “worried guy” at the end of the lesson: it seems like the amount of time spent moving at a given velocity must be considered, when calculating an average velocity of multiple travel segments of differing velocities and times. As an extreme example, if a car goes 60mph for 1 hour, then 30mph for 1 second, one can’t simply average 60 and 30 to get the average velocity. In the example problem, the fact that the worrier went 12.5 ft/min for 2 minutes seems to have been omitted from the calculation. I believe the correct solution should be 10+10+25 = 45 ft (total travel) / 1+2+1 = 4 minutes (total time) = 11.25 ft/min. Without rounding, the approach shown in the video yields an average velocity of 10.8333 ft/min. If I’m wrong, please help me understand my error!

  12. Aurora says:

    Actually you do… if you were to measure the exact velocity at every point (called instantaneous velocity) from here to there, add them up and divide by the number of points, then you’d get the same number.

  13. Sharon Hoerichs says:

    Why is that we name the overall velocity “average” but we don’t really average those numbers; we just add them up.

Have a question?

Tell us what you're thinking...

You must be logged in to post a comment.