Question: A gunshot is heard first by one observer and then $1.5$ seconds later by a second observer three kilometres away.
Assume that the gun and the ear levels of the observers are all in the same horizontal plane and that the speed of sound is $340$ $ms^{-1}$.
Choosing an appropriate coordinate system, write an equation that describes the possible positions of the gun, relative to the observers. Draw a diagram to illustrate your answer, indicating the possible positions of the gun relative to the two observers.
My attempt:
Suppose Observer 1 $(O_1)$ hears the gunshot $t_1$ seconds after the gunshot, and $(O_2)$ hears it $t_2$ seconds after.
Their distances from the gunshot are $340t_1$ and $340t_2$
$\therefore$ The differences between the distances is $340(t_2 - t_1)$ $= 340 \cdot 1.5 = 510$m
The differences between the two distances is a constant $510$m so this is a hyperbola, $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ , where $O_1$ and $O_2$ are at the foci and the gun is on the hyperbola.
The gun must be on $O_1's$ branch as hears it first.
Let the observers positions be $(\pm ae , 0)$
Then $2ae-(ae-a)$ = $ae+a $=$ 510m$
$a(e+1) = 510$
$a = \frac{510}{e+1}$
As $b^2 = a^2(e^2-1)$
$b^2 = \frac{510^2}{(e+1)^2}(e^2-1)$
$$ \therefore \frac{x^2(e+1)^2}{260100} - \frac{y^2(e+1)^2}{260100(e^2-1)} = 1 $$
where $e>1$ , $x<0$
This is my approach , I do not know if this is correct or not (my approach or solution)
I was wondering if you guys could show me a better way to approach the question and solve it thanks..
You’re correct that the source of the sound lies somewhere on an hyperbola with the two observers at the foci. You may have made things a bit messier than they need be by working with the eccentricity, though.
For an hyperbola in standard position with foci at $(\pm c,0)$, we have $c^2=a^2+b^2$, so its equation can be written as ${x^2\over a^2}-{y^2\over c^2-a^2}=1$. The difference in distance from a point on the hyperbola to the two foci is $2a$, which you can verify yourself by considering the points $(\pm a,0)$. In terms of the shot-spotting problem, we have $a = s\Delta t/2$, where $s$ is the speed of sound. For problems like these, I prefer to work in times instead of distances to keep the numbers small (the problem statement does leave the choice of coordinate system open). In that coordinate system, the equation becomes $${x^2\over(\Delta t/2)^2}-{y^2\over(c/s)^2-(\Delta t/2)^2}=1.$$