Chapter 27, Problem 2TS

If you know the rate of movement along a fault, the amount of offset over a period of time can be calculated. The basic relationship is

$Rate=\frac{distance}{time}$

Movement along the San Andreas Fault is about $3.5\text{cm}/\text{yr}$. If a fence were built across the fault in 2000, how far apart will the two sections of the new-broken fence be in 2010?

To determine

The distance between two sections of the new-broken fence.

Answer to Problem 2TS

The distance between two sections of the new-broken fence is $35\text{cm}$.

Explanation of Solution

Given info: The rate of movement along the San Andreas is $3.5\text{cm}/\text{yr}$ and time to cover the distance between two sections is $10\text{yr}$.

Write the expression for the distance between sections.

$d=n\times r$

Here,

$n$ is the time to cover the distance between two sections

$d$ is the distance between two sections

$r$ rate of movement of the San Andreas

Substitute $3.5\text{cm}/\text{yr}$ for $r$ and $10\text{yr}$ for $n$ in the above expression to get $d$.

$\begin{array}{c}d\left(10\text{yr}\right)\left(3.5\text{cm}/\text{yr}\right)\\ =35\text{cm}\end{array}$

Conclusion:

Therefore, the distance between two sections of the new-broken fence is $35\text{cm}$.