Question

What is the difference between average speed and root mean square speed, in particular for the situation when describing the velocity of particles?

Answer 1

\[Avg speed=\frac{ T.Distance }{ T.Time }\]

Answer 2

Probably nothing. The usual way of defining "average" speed in such a case is to use the root mean square velocity.

But you should be very careful. You said "speed' and not "velocity," and I have answered on that basis. Remember "speed" is the magnitude of velocity, which is a vector that has both magnitude and direction. "I am driving 60 MPH" is a statement of speed. "I am drivign 60 MPH west" or "I am driving 60 MPH east' are statements of velocity. You see the difference, I hope?

The reason it matters is that most of the time we deal with systems of particles in which the center of mass it at rest. For example, when we deal with 10 L of argon gas at STP, we normally asume the bottle holding the argon gas is at rest, and not (for example) speeding towards the Moon at 38,0000 MPH.

Under those circumstances the average VELOCITY of a particle is zero, because particles are equally likely to be going left as right, upwards as downwards, and so forth. (If the average particle were more likely to be going left than right, that could only mean that the entire sample of gas was moving to the left.)

However, the average SPEED is not zero, because the particles are generally moving quite fast.

An analogy may help: imagine a busy bridge into a city, like the George Washington Bridge into New York City. We can measure the average speed of the cars on it, and the average velocity. To measure the average speed, we just look at all the speedometers, write down the numbers, add, and divide by the number of cars. Let's say we get 45 MPH (which is optimistic, ha ha). To measure average velocity, however, we need to know the direction of each car, too. So we write down for Car #1 "48 MPH into the city" and for Car #2 "52 MPH out of the city" and so forth. Only then do we average, and what I hope you can see is that the motion of the cars out of the city will cancel some or all of the motion of the cars into the city, so that the average velocity might be quite different from the average speed.

For example, in the morning commute, we might find that the average velocity is 20 MPH into the city, because more cars are going in than out. In the evening commute hours, we might find the average velocity is 30 MPH out of the city, because more cars are leaving than entering. Finally, smack in the middle of the day, we might find the average velocity is zero, because just as many cars are going in as out, and traveling at about the same speed.

Answer 3

\[Vrms = \sqrt{\frac{ 3RT }{ M }}\]