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Using the same method, determine the speed for the remainder 18.11. Calculate the geothermal gradient in terms of °C/km by
of the age and distance values in the table. Results are show dividing the temperature change by the change in depth.
below.
T surface = 10°C __
T depth − T surface

6
_ T depth = 225°C geothermal gradient = depth − 0
1.8 × 10 cm
v 0.85 =

6
0.85 × 10 yr depth = 6.0 km __
225°C − 10°C

=
_ geothermal 6.0 km − 0 km
cm
= 2.1 gradient = ? 225 − 10 °C _
yr

6

_ = _
1.9 × 10 cm

v 1.0 = 6.0 − 0 km
6
1.0 × 10 yr 215 °C _
_
_ =
cm
= 1.9 6.0 km
yr
6
_ = 36 °C _
3.5 × 10 cm

v 1.8 = km
6
1.8 × 10 yr
_ 18.12. Since the volcanic arc is east of the trench, examination of
cm
= 1.9 Figure 18.15 shows the Nazca plate is subducting to the east.
yr
6
_
4.3 × 10 cm

v 2.0 = The calculation of slope is explained in appendix A. Th e y value
6
2.0 × 10 yr
is negative because it is a depth. In geoscience, slope is
_
cm
= 2.2 expressed in units of L/L, so the units are not canceled.
yr
Δy
_
Average the results. Δx = 300 km slope =
Δy = −100 km Δx
cm
cm
_
cm
cm _ _ cm _
_

−100 km

2.0 + 2.1 + 1.9 + 1.9 + 2.2 slope = ? _

____ = 300 km
yr
yr
yr
yr
yr
v average =

5 km

_ = −0.3 _
cm
= 2.0 km
yr
18.13. Since the volcanic arc is north-northwest of the trench,
18.10. Examine the table to determine in which direction each station
examination of Figure 18.16 shows the Pacific plate is
is moving and where they are positioned relative to one
subducting to the north-northwest. The y value is negative
another. Station 2 is to the north and east of station 1 because
because it is a depth.
its northing and easting coordinates are greater. In the case of
station 1, its motion is to the west because the easting data are Δx = 250 km _
Δy
slope =
decreasing while the northing data are not changing. In the Δy = −100 km Δx
case of station 2, its motion is to the east because the easting slope = ?
Convert the y value from km to m.
data are increasing while the northing data are not changing.
3
Determine the speed of each plate by using equation 2.1. _ )
1 × 10 m
−100 km (

1 km
Station 1 5
−1 × 10 m
d = 11.552 m − 12.000 m = −0.448 m _
5
−1 × 10 m

t = 2009 − 1995 = 14 yr slope = 250 km
v = ? m _
d _ = −400

v = km
t
_ 18.14 As Figure 18.18 shows, earthquake foci delineate the upper
−0.448 m

=
14 yr
surface of the subducting plate, so the depth and distance of an
−2 m _
= −3.2 × 10 earthquake from the trench can be used to determine the slope

yr
of the subducting plate. Calculate the slope based on each of
Station 2 the distance/depth pairs, then average the speeds. Th e y value is
d = 1,035.406 m − 1,035.000 m = +0.406 m negative because it is a depth.
t = 2009 − 1995 = 14 yr
v = ? Data Pair 1
d _
Δy
v = _
t Δx = 15.7 km slope =
_ Δx
+0.406 m

= Δy = −6.4 km
14 yr Convert the y value from km to m.
−2 m _
= +2.9 × 10 slope = ? 3

yr
1 × 1 0 m

−6.4 km ( _ )

Comparison of the motion of each station indicates the stations 1 km
3
are moving apart from one another; hence, they are situated on −6.4 × 1 0 m
a divergent plate boundary. _
3
−6.4 × 10 m

slope =
15.7 km
m _
= −410
km
682 APPENDIX E Solutions for Group A Parallel Exercises E-40