Understanding the Basic Concept of Average Speed

[serious7]

Hi guys, I'm having trouble understanding what average speed is all about. The equation that I know of calculating average speed is:

a(subscript av)=delta d/delta t.

 

Now is average speed the total distance travelled over total time as my book says it is? I'm kinda confused at that because how can the total distance travelled divided by the total time travelled give you the average speed. Isn't that the instantaneous speed? Suppose i'm observing a distance-time graph and the line is not straight but rather curved. If i follow the terminology from the book, all i would "need" to do to find the average speed is look at the total distance travelled and divide it by the total time travelled which should be the point at the end of the graph. How can that be the "average speed" of an object? Isn't it the instantaneous speed of an object at that particular moment?

 

If I didn't make any sense of my question then please tell me so I can clarify it.

Thanks.

[Sisyphus]

The definition in your book is correct. I'm here now, and an hour ago I was 50 miles away. By necessity, my average velocity during that hour was 50mph. I'm not quite sure what the source of you confusion is. What "particular moment" are you talking about?

[serious7]

OK how about this. How can I find the average speed from a distance-time graph if the line is not straight but rather curved?

[Sisyphus]

You find average velocity the way your books says to. Remember that you're finding the average velocity over a finite amount of time, i.e., between two points in time. It's just the difference between the distances divided by the time elapsed.

 

Suppose your equation is d=t^2, with distance in meters and time in seconds. At t=0, d=0 as well. At t=4, d=4^2=16. You've thus travelled 16 meters in 4 seconds, and so your average velocity is 16m/4s = 4m/s. Your instantaneous velocity is changing, from 0m/s to 8m/s.

[serious7]

So it is always the total distance divided by total time then. My next questions is this equation. Vav=d2-d1/t2-t1 If i were to find the average speed of an object from a distance-time graph, would d1 and t1 always be 0.0 km and 0.0 minutes? And when will it not be 0.0? And how did you get an instantaneous speed of 8 m/s in your example?

[Sisyphus]

Yes, it is always the total distance divided by the total time. d1 and t1 do not have to be zero. They can be anything. You just subtract d2-d1.

 

For example, in the same equation as before (d=t^2), suppose I want to find the average velocity between t=3 and t=4. At t=3, d=3^2=9. At t=4, d=4^2=16. Therefore, d1 would be 9, and d2 would be 16, so d2-d1=16-9=7. Similarly, t2-t1=4-3=1. So you've travelled 7 meters in 1 second, so your average velocity between t=3 and t=4 is 7m/s.

 

The 8m/s is just from the slope of the curve at t=4, representing the rate at which d is increasing compared to t at that particular point. It's found from differentiating the equation, which you'll learn in precalculus.

 

Oh, and BTW. You should really use the word "velocity" and not "speed," since they mean different things, and we're talking about velocity. Velocity cares about direction, and speed doesn't. For example, if I drive around a racetrack and end up back where I started, my average speed could be very fast, but my average velocity would be zero, because my position is the same at start and finish. This is because at opposite sides of the track, my velocities were in opposite directions and cancelled each other out.

[swansont]

Oh, and BTW. You should really use the word "velocity" and not "speed," since they mean different things, and we're talking about velocity. Velocity cares about direction, and speed doesn't.

 

Actually serious7 has been talking/asking about speed and has been using distance (scalar) and not displacement (vector). So speed is appropriate for that.