Astronomical distances are so large that using meters is cumbersome. For really large distances the light year is the best unit. A light year is the distance that light would travel in one year in a vacuum. Since the speed of light is fast, and a year is long, the light year is a pretty good unit for astronomy. One light year is about ten trillion meters as the following calculation shows.
v̅ = | Δs | ||
Δt | |||
Δs = | v̅Δt | = (3.0 × 10^{8} m/s)(365.25 × 24 × 3600 s) | |
Δs = | 9.46 × 10^{15} m | ||
Since both the speed of light and the year have exact defined values in the International System of Units, the light year can be stated with an unnecessarily large number of significant digits.
v̅ = | Δs | ||
Δt | |||
Δs = | v̅Δt | = (299,792,458 m/s)(365.25 × 24 × 3600 s) | |
Δs = | 9,460,730,472,580,800 m | ||
Some distances in light years are provided below.
Notice that no numbers are stated in this problem. When a numerical value is needed to solve a problem and that number is not given, it could mean one of several things.
In order to calculate speed, you will need distance and time. What distance does a point on the equator move in a convenient period of time? Well, I hope you know that the earth rotates once on its axis every day. You should also know how to calculate the length of a day in seconds. (A day is the period of the earth's rotation, for which an upper case T is the symbol.) During a day, a point on the earth's equator would have traveled a distance equal to the circumference of the earth. The radius of the earth is a number commonly found in textbooks and on reference tables. The problem can now be solved.
v̅ = | Δs | = | 2πr | = | 2π(6.4 × 10^{6} m) | = 470 m/s |
Δt | T | 24 × 3600 s |
That's about one-third greater than the typical speed of sound. An interesting problem to be dealt with later is that if the earth is spinning so rapidly, why then don't things on the equator fly off into space?
This problem is deceptively easy. Averaging is taught in elementary school, which makes this an elementary problem. Right?
6.0 km/h + 5.0 km/h | = 5.5 km/h |
2 | |
The Wrong Method of Averaging |
Wrong! Wrong! Wrong! Wrong! Wrong! You weren't paying attention in elementary school, were you? This is another example of how memorizing a procedure does not make you smarter (only less ignorant).
The add-and-divide method of averaging only works when averaging items of equal weight. The average age of the students in a classroom is the sum of their ages divided by the number of students only because each student is considered to have the same weight (a student, is a student, is a student, … ). In this problem, however, the two segments of the walk are significantly different. The second "half" was actually the majority of the walk. It carries more weight than the shorter first "half". Thus, the add-and-divide method won't work.
Let's return to our definition. Since speed is the rate of change of distance with time, we'll need both the distance traveled and the time it took to complete the walk. After we determine both of these numbers, the rest is easy.
Δt = | Δs | v̅ = | Δs | ||||||
v̅ | Δt | ||||||||
Δt_{1} = | 6.0 km | = 1.0 h | v̅ = | 6.0 km + 10 km | = | 16 km | |||
6.0 km/h | 1.0 h + 2.0 h | 3.0 h | |||||||
Δt_{2} = | 10 km | = 2.0 h | v̅ = | 5.3 km/h | |||||
5.0 km/h |
Look closely at the calculations on the right side. Notice that the formula contains Δ (delta) symbols and yet I added the distances in the numerator and the times in the denominator. That's because Δ doesn't mean difference, it means change. During the walk my position didn't change from 6.0 km to 10 km, it changed first by 6.0 km and then by 10 km for a total change of 16 km.
Velocity is the rate of change of displacement with respect to time. Velocity is a vector, which means the problem should be solved graphically. Draw an arrow pointing toward the top of the page (north). Label it 6 km. Draw another arrow to the left (west) starting from the previous one (arranged head to tail). Make it slightly longer and label it 10 km. Draw a third arrow starting on the tail of the first and ending on the head of the second. Since north and west are at right angles to one another, the resultant displacement is the hypotenuse of a right triangle. Use pythagorean theorem to find its magnitude and tangent to find its direction.
r = | √[(6.0 km)^{2} + (10 km)^{2}] | ||||
r = | 11.6619 … km | ||||
tan θ = | opposite | = | 10 km | ||
adjacent | 6.0 km | ||||
θ = | 59° on the west side of north | ||||
Divide displacement by time to get velocity.
v = | Δr | = | 11.6619 … km at 59° W of N |
Δt | 3.0 h | ||
v = | 3.9 km/h at 59° W of N |
This is an exercise in the factor label method. Write the value as a proper fraction and multiply by ratios equal to one with the intent of canceling the bad units and replacing them with the good ones. Everyone should know (or at least understand) that there are …
60 × 60 = 3600 seconds
in an hour. Many Americans who are fans of track and field know that four laps around a 400 m outdoor track is almost one mile.
1 mile ≈ 4 × 400 m ≈ 1600 meters
More precisely … actually, most precisely … actually, exactly by definition …
1 mile = 1609.344 meters
60 miles | 1609.344 m | 1 km | ≈ 96.56 km/h | ||
1 hour | 1 mile | 1000 m | |||
60 miles | 1609.344 m | 1 hour | ≈ 26.82 m/s | ||
1 hour | 1 mile | 3600 seconds | |||