Chapter 10, Problem 81RE

To determine

How far north should the site of the second explosion be?

Answer to Problem 81RE

  $\boxed{450feet}$

Explanation of Solution

Given information:

In a test of their recording devices, a team of seismologists positioned two of the devices $2000feet$ apart, with the device at point $A$ to the west of the device at point $B$ . At a point between the devices and $200feet$ 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:

Given that the measured time difference for the first detonation is equal to that the second detonation, find the time difference of the first detonation.

Let $t_1$ be the time required for the sound to reach the point $B$ and $t_2$ be the time required for the sound to reach the point $A$ .we know that the speed of sound is $1100$ feet per second.

Since $B$ is $200feet$ away from the point of the first detonation, $A$ $2000-200=1800feet$ away from the point of the first detonation. So, the time taken for the sound to reach $B$ is,

  $\begin{array}{l}{t}_1=\frac{dis\tan ce}{speed}\\ =\frac{200}{1100}\\ =\frac{2}{11}\end{array}$

So, the taken from the sound to reach $A$ is,

  $\begin{array}{l}{t}_2=\frac{dis\tan ce}{speed}\\ =\frac{1800}{1100}\\ =\frac{18}{11}\end{array}$

So, the measured time difference, $t_2-t_{1}is\frac{18}{11}-\frac{2}{11}=\frac{16}{11}\sec .$

Since the time difference for the first detonation is equal to that of the second detonation, the time difference for the second detonation will also be $\frac{16}{11}\sec .$

Using the difference, find the difference between the distance of $BtoA$ from the second point of detonation, which lies directly above $B$ . Let that point be $\left(x,y\right)$ and $d_1$ and $d_2$ be the distance of $B$ and $A$ from $\left(x,y\right)$ . Then we have $\left(d_2-d_1\right)=\left(t_2-t_1\right)1100$ .

Substitute $\frac{16}{11}$ for $\left(t_2-t_1\right)$ in $\left(d_2-d_1\right)=\left(t_2-t_1\right)1100$

  $\begin{array}{l}\left(d_2-d_1\right)=\left(\frac{16}{11}\right)1100\\ =1600\end{array}$

Since the difference of the distance is $1600feet$ assume that the point $\left(x,y\right)$ lies on a hyperbola whose foci are at $AandB$ . We have $d\left(F_1,P\right)-d\left(F_2,P\right)=2aor2a=1600$ .

  $\begin{array}{l}\frac{2a}{2}=\frac{1600}{2}\\ a=800\end{array}$

Since the foci are $2000feet$ apart, the points are $\left(1000,0\right)$ and $\left(-1000,0\right)$ . So

  $\begin{array}{l}2c=2000\\ c=1000\end{array}$

We have $b^2=c^2-a^2$ put the values,

  $\begin{array}{l}{b}^2={1000}^2-{800}^2\\ =360000\\ \sqrt{b^2}=\sqrt{360000}\\ b=600\end{array}$

Replace $a,b$ in the standard parabola equation,

  $\begin{array}{l}\frac{x^2}{{800}^2}-\frac{y^2}{{600}^2}=1\\ \frac{{1000}^2}{{800}^2}-\frac{y^2}{{600}^2}=1\left[\therefore x=1000\right]\\ \frac{y^2}{{600}^2}=\frac{{1000}^2}{{800}^2}-1\\ \frac{y^2}{{600}^2}=\frac{{1000}^2-{800}^2}{{800}^2}\\ \frac{y^2}{{600}^2}=\frac{{600}^2}{{800}^2}\\ \sqrt{\frac{y^2}{{600}^2}}=\sqrt{\frac{{600}^2}{{800}^2}}\\ \frac{y}{600}=\frac{600}{800}\\ y=\frac{600}{800}\times 600\\ y=450\end{array}$

Hence the second detonation should be $\boxed{450feet}$ to the north of $B$ .