practice problem 1
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.
Start with the definition of speed and solve it for distance.
|∆s = c∆t||⇐||
Numbers in, answer out.
|Δs = cΔt
Δs = (3.0 × 108 m/s)
Δs = 9.46 × 1015 m
Since both the speed of light and the year have exactly defined values in the International System of Units, the light year can be stated with an unnecessarily large number of significant digits.
|Δs = cΔt
Δs = (299,792,458 m/s)
Δs = 9,460,730,472,580,800 m
Some distances in light years are provided below.
- The distance to Proxima Centauri (the star nearest the sun) is 4.3 light years.
- The diameter of the Milky Way (a collection of stars that includes the sun and all the stars visible to the naked eye) is about 100,000 light years.
- The distance to Andromeda (the nearest galaxy outside the Milky Way) is about 2 million light years.
- The distance to the edge of the universe (the observable part of it) is 13.77 billion light years.
practice problem 2
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.
- Look it up! It may appear somewhere in the textbook you are using — on the inside covers, in an appendix, or in the text of the chapter you are currently working on. It may be found in the reference table that some teachers distribute. Standardized exams usually also have their own reference table.
- Know it! Some numbers are numbers that you should just know. In this problem, there is one relevant number that nearly everyone knows. You may also be expected to memorize certain numbers by an instructor or professor.
- Calculate it! Maybe there's a way to find the number you need to know using other numbers given in the problem.
- Forget about it! Maybe you don't really need the number you think you do. Maybe you are on the wrong track. Especially under test conditions, it is highly unlikely that you could be asked a question that requires a numerical value that you can not find, do not know, or can not calculate. Perhaps there is another method to solve this problem.
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 = 470 m/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?
practice problem 3
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 Δt1 = 6.0 km = 1.0 h 6.0 km/h Δt2 = 10 km = 2.0 h 5.0 km/h v̅ = Δs Δt v̅ = 6.0 km + 10 km 1.0 h + 2.0 h v̅ = 5.3 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 Δt v = 11.6619… km at 59° W of N 3.0 h v = 3.9 km/h at 59° W of N
practice problem 4
Everyone should know (or at least realize after a bit of thought) that there are…
∆t = 60 × 60 = 3,600 s
…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.
∆s = 1 mile ≈ 4 × 400 m = 1,600 m = 1.6 km
More precisely… actually, most precisely… actually, exactly by definition…
∆s = 1 mile = 1,609.344 m = 1.609344 km
The first answer…
v = ∆s ∆t v = 60 miles 1 hour v = 60(1.609344 km) 1 h v = 96.6 km/h
For comparison, the speed limit on many of Canada's highways is 100 km/h.
The second answer…
v = ∆s ∆t v = 60 miles 1 hour v = 60(1,609.344 m) 3,600 s v = 26.8 m/s
You should note that the number with International units is a little bit less than half the value of the number with British-American units. I've gotten used to mph, but I have to be conversant in m/s for my job. A good rule of thumb for comparing speeds is to…
divide by 2 and subtract a little
when converting from mph to m/s
multiply by 2 and add a little
when converting from m/s to mph.