Dot Physics

It has been windy here lately. Sometimes I think that is an ok thing. You see, when I ride my bike to work I am probably going to have the wind at my back for one of directions. It is great feeling like Lance Armstrong because of the boost you get from the wind. With a good wind at my back, I can almost keep up with the traffic (I would keep up if they went the 25 mph speed limit).

Of course, with a great boost comes a great drag. When I ride into the wind, I feel weak. I pedal as fast as I can and cars just whiz right by like I am standing still. When you are in a car, you don’t really notice the wind. You notice it on a bike as though it were blowing in your face (because it is).

Yesterday it was windy. It seemed the god of the wind (Hermes? I don’t know who you are, god of the wind – maybe that is why you smote me) was always in my face. It was no fun. When life gives you lemons, blog about it.

So, here is one of my favorite easy physics problems. Suppose I bike 2 miles to work. On the way there, the wind gods like me and I can go 20 mph. On the way back, I only go 10 mph. What is my average speed for the round trip?

First, notice that I said “average speed” and not “average velocity”. The convention is that average velocity is defined as:

i-017c0320fafd1a002e749d0472ed4fb9-avg-vel-1.jpg

Where the vector r is the position of the object. If I take a round trip, my starting and ending position vectors are the same thing. This would make the average velocity zero (zero vector). Average speed, on the other hand, can be calculated as:

i-0d920c2e44054ed88ec238f9a52bbd45-avgspeed-1.jpg

Where s is the total distance along the path (not zero for a round trip).

Ok, I want to calculate average speed. For this case, let me pretend that I am going all in one direction. So the problem would be that I travel 20 mph for 2 miles and then 10 mph for 2 miles. Now I can call my position along the x-axis.

Let me go ahead and give you the wrong answer. The wrong answer is that the average speed is (20+10)/2 = 15 mph. This is incorrect. This would be correct if someone went 20 mph for 10 min then 10 min for 10 min. Let me just do this the long way.

i-d8d5443e4bfaa4f2244f9a2fc58febcb-avgsppedpic-1.jpg

To find the average speed, I need the total distance (got that) and divide by the total time (don’t have). I can find the time for the first part of the trip and for the second part. To make this more generic, I am going to call the first speed v1 and the second speed v2. I will let the starting position be x = 0 miles. The distance where the bike changes speed will be x2. The final distance will be x3. The time for the first part of the trip will be: (note, I will am using v without a vector symbol to denote speed)

i-2a63b0295cb9a29ac9e7bc9c550b9a81-t-1213.jpg

Now, to do the same thing for the second part of the trip:

i-302030113e1eb2c5564f22a9ca5527d7-time-2.jpg

The average speed is now the total distance divided by the total time. The total time is:

i-d71a1bd671984f5248fa78715953bf98-totaltime.jpg

The total distance is x3, so the average speed is:

i-7ce74e1aecbc049c6fd3a8ba18398550-avgtotalspeed.jpg

Before I plug in the numbers – there is on check I can make. Are the units correct? On the top, I have position times v2 (units). On the bottom are position and velocity units. These will cancel to just speed units – so that is good. Also, what if v1= v2? In this case, the average speed should be v1. If you plug into the equation above, that is indeed what you get. Now to plug in my numbers.

i-5bf13384d8b6b262ef026395cc15d18f-ansdfw-3r.jpg

This is lower than the 15 mph as it should be. The bike was traveling at a lower speed for a longer time than it was at the greater speed. What if I wanted to find the average speed after going v1 for a time t1 and then v2 for a time t2? This is slightly easier. The total time is just t1 + t2. However, I need to find the distances:

i-413f19ed9b630d1595e0b44dee501c35-timeave-1.jpg

Now for the average speed:

i-5a8320c14c5fabfc2f9e3fed916a136c-weightedv.jpg

This is simply the time-weighted average of the two speeds. Ok – I was afraid of an angry god, so I looked it up. Aeolus is the king of the winds.