The following pages contain numerical word problems
presented in lecture. There are two steps to solving word problems :
1. translate the wording into a numeric equation;
What are
the variables, constants (for example, distance (d), rate ( r)
2. solve the equation!
This may first involve simplifying the equation
(combining like terms, etc)
1. You
are in the Pacific Ocean after having sailed from Santa Cruz 4 days earlier. It
is noon (12:00) on the ship (local time), but a clock that keeps Santa Cruz
time is at 3 PM (15:00). How many degrees
of longitude have you traveled relative to Santa Cruz?
Solution:
Recall: 360¡ longitude / 24 hr = 15¡/hr
15:00-12:00=3
hr
3 hr x 15¡/hr = 45¡
2. You are on a
survey ship in the Atlantic Ocean measuring water depth using sonar. The sonar records two way travel time
(from ship to seafloor and back) of 6 seconds. Assuming that the average
velocity of sound in seawater is 1.5 km/sec, determine the water depth in
meters.
Solution:
d = r x t, r=1.5 km/sec, t=6 sec
d =1.5 km/sec x 6 sec = 9 km
this is the two-way (roundtrip) travel time /distance
9 km/2 = 4.5 km
4.5 km x 1000 m/km = 4500 m
3. Computing distance from an earthquake epicenter is easy if you can record differences in the arrival times of primary and secondary waves. You must also know the mean velocity of the waves. First derive an equation that can be used for this purpose.
What do we know?
VP = 6.3 km/sec, VS = 3.6 km/sec
¥
Difference
in arrival time (ÆT) between
p and s waves = TS - T P
Time it takes p and s waves to travel the same distance (D)
¥
TP = D/6.3
, TS = D/3.6
ÆT between the
P- waves and S-waves is:
¥
ÆT = TS -
T P = D/3.6 Ð D/6.3 = 1.75D/6.3 - D/6.3
or
¥
ÆT = 0.75D/6.3
The distance D from the earthquake epicenter to seismic
station is:
¥
D = 8.4
(km/sec) ÆT
8.4
represents the conversion factor.
This can be simply computed by dividing the product of the velocities by
the sums.
VP x VS / VP - VS = (6.3 x 3.6)/(6.3-3.6) = 8.4
4. Given the equation above, if S-wave arrives 20 seconds after the primary wave, what is the distance to the earthquake?
Answer: 168 km.