MAE 6328
Assignments due Mar 21, 2007
Alice Baeza
Spring 2007
1. For each of the following questions, decide whether the question is asking for the distance or the
magnitude of the displacement. Describe your reasoning.
a. How far do you walk each day? Distance – we can measures from where we start to were we
end.
b. How far is it from Toronto to Mexico City by airplane? Displacement – it is described by
magnitude because the plane will travel in the most possible straight line.
c. How far is it from Toronto to Mexico City by car? Distance – you cannot go on a straight line
in a car from one place to another. Each route has different components in direction
therefore we are measuring the distance from one place to another.
d. How far is the earth from the moon? Displacement – The orbit changes over the course of the
year so the distance from the Moon to Earth roughly ranges from 357,000 km to 407,000
km.
e. How long is a standard sheet of plywood? Displacement – Distance - The measurement is a
straight line.
f. How long is the coastline of Britain? Distance – you cannot measure it in a straight line and
the coastline is infinitely long
g. How long is the Nile?- Distance –From Nile Source to Mouth is 6695km (4184 miles).
h. How long is the track on a compact disc? – Distance –because we are traveling in a spiral
motion. The total length of the track on a CD is almost 3 miles (~4.5 km).
i. How long did it take you to get here? Distance – Time is measured in terms of distance.
j. How many frequent flyer miles do you have? None – It is not distance traveled, nor time is a
variable to consider. There is no displacement from a starting and ending point either.
k. What is the radius of the earth? Distance -Equatorial radius 6378.1(km). The radius is a
straight line from the center of the earth.
2. Use the following conversion factors:
1 mi = 5280 ft 1 ft = 12 in 1 in = 2.54 cm
1 h = 60 min
1 min = 60 s
1 m = 100 cm
1 km = 1000 m
A. to convert 60 mi/h (mph) to km/hr.
= (60/hr) (5280ft/1mi) (12in/1ft) (2.54cm/1in) (1m/100cm) (1Km/1000m)
=((60) (5280) ( 12) (2.54)Km) / ((100) ( 1000) m)
= 9656064/100,000
96.6Km/Hr
B. to convert 60 mi/h (mph) to ft/s.
= (60/hr) (5280ft/1mi) (1hr/60min) (1min/60sec)
= ((60) (5280) (1) (1)ft) / ((60) ( 60) sec))
= 316800/3600
= 88ft/sec
MAE 6328
Assignments due Mar 21, 2007
Alice Baeza
Spring 2007
C. to convert 60 mi/h (mph) to m/s.
= (60/hr) (5280ft/1mi) (1hr/60min) (1min/60sec) (12in/1ft) (2.54cm/1in)(1m/100cm)
= ((60) (5280) (1) (1)(12)(2.54)(1)m) / ((60) ( 60) (100) sec))
= 9656064/360000
= 26.82m/sec
3. In an experiment at James Cook University in Australia, a researcher put the larvae of tropical fish in
a special tank to measure their swimming speeds. The tank generates an adjustable current that the fish
must swim against. The most proficient swimmer was a surgeonfish larva that maintained a 13.5 cm/s
swim for an equivalent distance of 94 km without a rest. For how long was the larva swimming?
(13.5cm/s) (1m/100cm) (1km/1000m) (60sec/1min) (60min/1hr)
= (13.5)(60)(60) / (100)(1000)
= 48600/100000
=.486Km/ hrs
The number of hours to travel 94 Km is 94/.486 = 193.42 hrs.
4. Complete your own version of the distance – speed GeoGebra construction that was used in class and
is shown to the right. Suggest other distance functions s(t) that will provide a good illustration.