Chapter 10, Problem 41RE

To determine

To find: How far north should the site of the second explosion be chosen so that the measured time difference recorded by the devices for the second detonation is the same as that recorded for the first detonation?

Answer to Problem 41RE

450 feet due north.

Explanation of Solution

Given:

In a test of their recording devices, a team of seismologists positioned two of the devices 2000 feet apart, with the device at point $A$ to the west of the device at point $B$. At a point between the devices and 200 feet from point $B$, a small amount of explosive was detonated and a note made of the time at which the sound reached each device. A second explosion is to be carried out at a point directly north of point $B$. How far north should the site of the second explosion be chosen so that the measured time difference recorded by the devices for the second detonation is the same as that recorded for the first detonation?

Calculation:

Let the set of all points where an explosion takes place with the same time difference as that for first detonation would form a hyperbola with $A$ and $B$ as foci.

Since $A$ and $B$ are foci, we have, $2c=2000$ and $c=1000$.

From the figure above, $\left(a,0\right)$ is on the transverse axis. So, it must be the vertex of the hyperbola.

The vertex is 200 feet from $B$. We have $a=800$.

$b^2=c^2-a^2={1000}^2-{800}^2$

$b=600$

The equation of hyperbola is

$\frac{x^2}{640,000}-\frac{y^2}{360,000}=1$

The point $\left(1000,y\right)$ lies on the hyperbola. Substituting in the equation of hyperbola, we get,

$\frac{{1000}^2}{640,000}-\frac{y^2}{360,000}=1$

$y=450$ feet

The second explosion should be 450feet due north so that measured time difference recorded by the devices for the second detonation is the same as that recorded for the first detonation.