Hello sorry lads,
For the given questions, please refer to the following diagram (all linear measures are in km):
https://www.physicsforums.com/attachments/848._xfImport
The plane is at point $P$ and the radar station is at point $R$.
a) Express $D$ as a function of $S$, and $S$ as a function of $t$.
To express $D$ as a function of $S$, we may observe that the Pythagorean theorem allows us to write:
$$S^2+4^2=D^2$$
Now, taking the positive root since lienar measures are taking to be non-negative, we have:
$$D(S)=\sqrt{S^2+4^2}$$
To express $S$ as a function of $t$, we may use the fact that for constant speed, distance is speed times time elapsed. For now, we will let the positive constant $v$ represent the speed of the plane, and we have:
$$S(t)=vt$$
b) Use composition to express the distance between the plane and the radar station as a function of time.
$$D(t)=D(S(t))=\sqrt{(vt)^2+4^2}$$
As you can see, we merely need to substitute $S(t)$ for $S$ in $D(S)$ to get $D(t)$
c) How far from the station is the plane 10 minutes later?
Since time is in hours (to be consistent with $v$ which is in kph), this means:
$$t=10\text{ min}\cdot\frac{1\text{ hr}}{60\text{ min}}=\frac{1}{6}\,\text{hr}$$
Using the given value of $$v=560\,\frac{\text{km}}{\text{hr}}$$
we then find:
$$D\left(\frac{1}{6}\,\text{hr} \right)=\sqrt{\left(560\,\frac{\text{km}}{\text{hr}}\cdot\frac{1}{6}\,\text{hr} \right)^2+(4\text{ km})^2}=\sqrt{\left(\frac{280}{3} \right)^2+4^2}\text{ km}=\frac{4}{3}\sqrt{4909}\,\text{km}\approx93.41900829655124\text{ km}$$
To sorry lads and any other guests viewing this topic, I invite and encourage you to post other algebra questions here in our http://www.mathhelpboards.com/f2/ forum.
Best Regards,
Mark.
