Q8PE - College Physics (Urone)

Question

Land west of the San Andreas fault in southern California is moving at an average velocity of about $\mathbf{6}\mathbf{cm}/\mathbf{y}$ northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is $\mathbf{590}\mathbf{km}$ northwest, assuming the motion remains constant?

Step-by-Step Solution

Answer

$\mathbf{9}.\mathbf{83}\mathbf{x}{\mathbf{10}}^6\mathbf{y}$

Step 1: Given Data

• Speed of the drift  $\mathbf{=}\mathbf{6}\mathbf{\text{\hspace{0.17em}cm/y}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{06}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\text{\hspace{0.17em}km/y}}$.
• Distance of the drift  $=\mathbf{590}\mathbf{km}$ .

Step 2: Speed of the drift

How much distance is traveled in a given time can be used to determine the speed.

Speed does not need a direction to express it. It is a scalar quantity.

 $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$

 

Substituting values in the above expression, we get:

$\begin{array}{l}\mathrm{T}=\frac{590\text{\hspace{0.17em}km}}{0.06\times {10}^{-3}\text{\hspace{0.17em}km/y}}\\ \mathrm{T}=9,833,333.33\mathrm{y}\end{array}$

Hence in $\mathbf{9}.\mathbf{83}\mathbf{x}{\mathbf{10}}^6\mathbf{y}$ , they will be $\mathbf{590}\mathbf{km}$  apart from each other.