Kinematics Notes

Average Speed

[Chapter 2 Objectives]

BHS -> Mr. Stanbrough -> Physics -> Mechanics -> Kinematics -> this page

The average speed of an object tells you the (average) rate at which it covers distance. If a car's average speed is 65 miles per hour, this means that the car's position will change (on the average) by 65 miles each hour.

Average speed is a *rate*. In kinematics,
a rate is always a quantity divided by the time taken to get that
quantity (the *elapsed time*). Since average
speed is the rate position changes, average speed = distance
traveled/time taken.

Example:

A car travels between 2 towns 60 miles apart in 2 hours. What is
its average speed?

average speed = distance/time Therefore, the average speed of the car is 60 miles/2 hours = 30 miles/hour.

Example:

If a person can walk with an average speed of 2 meters/second, how
far will they walk in 4 minutes?

There are 60 seconds in 1 minute, so there are 4 (60 seconds) =
240 seconds in 4 minutes. Also, if average speed = distance/time,
then distance = (average speed)(time). Therefore, the distance the
person moves is (2 m/s)(240 s) = 480 meters.

Since average speed is always calculated as a distance (length) divided by a time, the units of average speed are always a distance unit divided by a time unit. Common units of speed are meters/second (abbreviated m/s), centimeters/second (cm/s), kilometers/hour (km/hr), miles/hour (mi/hr - try to avoid the common abbreviation mph), and many others.

Example:

Which of the following could be a speed measurement?

- 2.5 meters
- 2.5 seconds/meter
- 2.5 meters/second
- 2.5 meters/second/second

Only 2.5 meters/second could be a speed measurement. Speed always has units of a distance (length) unit divided by a time unit.

Which Distance?

Farmer Jones drives 6 miles down a straight road. She turns around and drives 4 miles back. What was her average speed for this trip if it took 1 hour? |

Your answer to this problem depends on your interpretation of "distance traveled". You could say:

- The
**total distance**traveled by Farmer Jones is 10 miles. Therefore her average speed is 10 mi/hr. - The
**net distance**traveled by Farmer Jones is 2 miles. Therefore, her average speed is 2 mi/hr.

There are good reasons to use either interpretation - it's mostly
a matter of preference. We will interpret "distance traveled" to be
**net distance (**also called** displacement).
**Farmer Jones' average speed was 2 mi/hr.

NOTE:Different texts may adopt other conventions! In fact, our AP Physics text uses total distance to calculate speed, but net distance to calculate velocity. Use caution here!

The Perils of Averaging Averages

Here is an interesting problem:

Susie has planned a trip to a city 60 miles away. She wishes to have an average speed of 60 miles/hour for the trip. Due to a traffic jam, however, she only has an average speed of 30 miles/hour for the first 30 miles. How fast does she need to go for the remaining 30 miles so that her average speed is 60 miles/hour for the whole trip?

Most likely you thought "Oh, 90 miles/hour - since the average of 30 and 90 is 60! Boy, this is easy!"

Unfortunately, however, the answer is **not** 90
miles/hour. Here's why: You know that average velocity =
distance/time (v = d/t). In order to have an average speed of 60
miles/hour over a distance of 60 miles, you must complete the trip in
1 hour:

But Susie has already taken an hour (it takes 1 hour to go 30
miles with an average speed of 30 miles/hour) - and she is only half
way! It is **impossible** for her to complete the trip
with an average speed of 60 miles/hour! She would have to go
infinitely fast!

Notice that it would take 1/3 of an hour to cover the last 30
miles at 90 miles/hour. The total time for her trip would be 1.33
hours, and her average speed would be:

Try this calculation for *any* speed for the second half of
the trip - the average speed for the whole trip *cannot ever be 60
miles/hour!* The moral of the story: **Don't average
averages!**

This would be a good time to do the Measuring Speed Activity, in which you:

- determine some average speeds by measuring distances and times, and
- determine an unknown distance by measuring time to cover the distance at a known speed

[Chapter 2 Objectives] BHS -> Mr. Stanbrough -> Physics -> Mechanics -> Kinematics -> this page

last update November 22, 2005 by JL Stanbrough