As you might have suspected
from the last lecture, its not easy describing the motion of the Earth through
space if you are on it, as opposed to looking down from a satellite or space
station. The task gets even harder if you don't understand **circular motion**,
**acceleration**, **velocity**, **speed**, and **inertia** as was
the case in the time of Aristotle. Without such concepts, it was difficult,
if not impossible, to relate the motion of ordinary projectiles to the motion
of celestial bodies such as the Earth.

"Wait just a sixtieth of an hour glass," you might say. "How could Anyone not understand speed? And what's so tricky about velocity, and acceleration?" A modern day Aristotle would have no problems with such ideas, of course. Like you, he would be accustomed to zipping from one place to another in cars and airplanes, with dials and digits informing him of his distance, time, and speed……..

Speed, admittedly , never was particularly troublesome. It can be defined one of two ways:

- What the speedometer says
- A distance divided by the time it took to travel that distance.

Now consider the following problem:

You start off from home and drive to Las Positas College. You left home at 2 PM, and arrived at Las Positas at 2:20 PM. Your odometer says that you have driven 4 miles. What was your speed?

You might be tempted to say, "Well, speed is distance divided by time. I traveled 4 miles in 20 minutes so my speed is

4 miles/20 minutes =

4 miles/(1/3) hour =

**12 miles/
hour**

Great! But if you glanced at your speedometer during the drive, chances are that it wouldn't have said 12 miles/hour. Most of the time, you were going faster than that, say 30 miles/ hour. But some of the time, you were stopped dead in tracks, waiting for the light to change or a train to go by. What we need here is two definitions of speed: Average speed, and instantaneous speed.

**Average
speed** = **total distance traveled/ total time of travel**

(12 miles/hr in this case)

In the best traditions of physics, you may ask, "well how does the speedometer define speed?" The answer is that it is defining instantaneous speed as the average speed over a very small time interval.

In other words…….

Instantaneous speed is a small distance divided by a small time interval. The distance your car travels when the wheel turns around once, for example, divided by the time it takes the wheel to turn. A better measure of instantaneous speed would be the distance your car travels when the wheel turns through half a turn (half the circumference of the wheel) divided by the time it takes to rotate through half a turn. Or you could use a quarter turn, or an eighth of a turn, or….I think you get the picture. The smaller the distance interval and time interval, the more accurate your measure of instantaneous speed.

While we are at it, lets
define Average and instantaneous **velocity**.

**Average velocity = (total
displacement) / (time of travel)**

**Instantaneous velocity
= What your speedometer says and what direction you are going at that
instant.**

Oh, Oh…lots of apparently new concepts at once…but don't worry, they aren't that new…first of all:

What is displacement? Image you walk from one corner of your living room to the diagonally opposite corner. You could do this by walking straight from one corner to another, or you could follow the walls, or you could zig zag and pirouette about until you got there. In all cases, the displacement would be the same: the straight line distance from where you started to where you ended up.

Notice that the displacement, together with the words "start" and "end" define a direction (let North be towards the top of the page, and East be to your right). If the length of the displacement is 30 feet, and the time of travel is 10 seconds, then your average velocity is:

**V** = 30 feet/second
**North-East**.

Next, consider a very small piece of the path (the shaded section)

As shown by the arrows at right above, you were headed in different directions at different times. A speedometer attached to your belt would tell you your speed. A compass could tell you what direction you were headed. Both those bits of information define your instantaneous velocity.

Question 1: What is the relation between instantaneous velocity and average velocity?

Let see if you can put it all together with this problem:

Question 2: Marty the mountain man, and Jake the dog hike a trail in the Sierra Nevada Mountains.

They hike from the trailhead to Cherry Lake, a straight line distance of 5 miles due east. Jake, being a happy dog, runs back and forth so that he actually covers a total distance of 20 miles by the time they both arrive at Cherry Lake 2 hours later.

- What was Marty's average speed?
- What was Jake's average speed?
- What was Marty's average velocity?
- What was Jake's average velocity?

Now this may not seem like the most exciting concepts you have ever seen, so let me try to wow you with some speeds (I'm sneaking some units in here as well):

Continental
drift: 2 mm/year (1mm = one millimeter = 10^{-3 }meters)

Walking speed: 3 miles/hour = 4.4 feet/second = 1.3 m/s

(1m = 1 meter, s = seconds)

Driving speed: 60 miles/hr = 100 km/hr = 88 ft/s = 26 m/s

Orbital speed (near Earth orbit) = 8 km/s = 17,000 miles/hr

Earth's speed around the sun = 30 km/s

Speed of light = 300,000 km/s = 186,000 miles/s.

As you can see, there is quite a range of speeds in somewhat familiar phenomena. While there is no limit to how slowly something can travel, there is an upper limit: the speed of light. We will discuss why this is much later. In the mean time, lets use the definition of average speed to explore a familiar relationship between speed and distance:

Total Distance = average speed x time.

d = vt

Without knowing it, you probably use this relationship every day. If you travel at sixty miles per hour for two hours, you travel 120 miles. Note that the 120 miles, in this case is the total distance you have traveled (i.e. the curves and pirouettes, etc). A similar relation is:

Displacement = average velocity x time interval

(Dd = |**wave**|
Dt)

Using the example of Marty and Jake above….

Jake has
an average velocity of 2.5 miles/hr due east and travels for 2 hours. His **displacement**
is 5 miles to the east.

You are also accustomed to using the relationship d = vt to find time. If you have an average speed of 30 miles/hr and you travel 120 miles, it takes a time of 4 hours. In other words, you did the following calculation:

…d = vt, or t = d/v = (120 miles)/(30 miles/hr) = 4 hr.

Next comes acceleration: Acceleration defined as the change in instantaneous velocity/ change in time.

…a = Dv/Dt

This means that if your velocity changes from 1 meter/s to 2 meters/s in a time of 1 second, then your acceleration is:

(2m/s - 1m/s)/1
s = 1 m/s/s = 1 m/s^{2}

Another way of looking at
this is that you gain velocity of one meter per second each second. Something
dropped near the earth accelerates downward with an acceleration of about 10
m/s^{2} (actually 9.8m/s^{2} or 32 feet/s^{2}). This
means that an object that isn't moving when you let go of it, will have a speed
of about 10 m/s after one second, 20 m/s after two seconds, 30 m/s after 3 seconds
and so on.

Here is a little "Free Fall" table to help you with these numbers:

Time (seconds) |
Speed (meters/s) |
Distance (meters) |

0 |
0 |
0 |

1 |
10 |
5 |

2 |
20 |
20 |

3 |
30 |
45 |

4 |
40 |
80 |

If you are following along, then the speed numbers look pretty good, and straight forward. A dropped object gains 10m/s in velocity each second. But why are the distance numbers so strange? The answer is that the object is speeding up as it falls, so it falls larger and larger distances each second. For instance, the average velocity of the object over the first second is 5 m/s (it starts off with a speed of 0 and ends with a speed of 10 m/s).

Thus in the first second it travels a distance of 5 m. In the second second, its average speed is halfway between 10 and 20 meters/s or 15 meters per second. Thus in that second it travels 15 meters, for a total distance of 15 plus 5. Well, you can perform this type of calculation indefinitely, or you can use this simple formula:

….h = 1/2 g t^{2}
where g is the acceleration of gravity (10 m/s^{2}), and t is the elapsed
time after the object falls. Try plugging in a few times and see if you values
agree with the above distances.

If you find all this too confusing, don't worry. Your book does a good job explaining acceleration and free fall, and I won't ask any questions on a test that require you to do more than the simplest calculation (and you will have plenty of practice for that).

Answer Q1: instantaneous velocity is the ratio of a very small displacement to a very small time interval. The smaller the displacements, the more accurate!