Welcome to our community

MarkFL

Staff member
Here is the question:

HELP WITH THIS FUNCTIONS QUESTION!!!!!!!!!!!!?

An airplane passes directly over a radar station at time t = 0. The plane maintains an altitude of 4km and is flying at a speed of 560km/hr. Let D represent the distance from the radar station to the plane, and let S represent the horizontal distance traveled by the plane since it passed over the radar station.

---

a) Express D as a function of S, and S as a function of T
b) Use composition to express the distance between the plane and the radar station as a function of time.
c) How far from the station is the plane 10 minutes later?
Here is a link to the question:

HELP WITH THIS FUNCTIONS QUESTION!!!!!!!!!!!!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

MarkFL

Staff member

For the given questions, please refer to the following diagram (all linear measures are in km): 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:

$$\displaystyle S^2+4^2=D^2$$

Now, taking the positive root since lienar measures are taking to be non-negative, we have:

$$\displaystyle D(S)=\sqrt{S^2+4^2}$$

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:

$$\displaystyle S(t)=vt$$

b) Use composition to express the distance between the plane and the radar station as a function of time.

$$\displaystyle D(t)=D(S(t))=\sqrt{(vt)^2+4^2}$$

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:

$$\displaystyle t=10\text{ min}\cdot\frac{1\text{ hr}}{60\text{ min}}=\frac{1}{6}\,\text{hr}$$

Using the given value of $$\displaystyle v=560\,\frac{\text{km}}{\text{hr}}$$

we then find:

$$\displaystyle D\left(\frac{1}{6}\,\text{hr} \right)=\sqrt{\left(560\,\frac{\text{km}}{\text{hr}}\cdot\frac{1}{6}\,\text{hr} \right)^2+(4\text{ km})^2}=\sqrt{\left(\frac{280}{3} \right)^2+4^2}\text{ km}=\frac{4}{3}\sqrt{4909}\,\text{km}\approx93.41900829655124\text{ km}$$

To sorry lads and any other guests viewing this topic, I invite and encourage you to post other algebra questions here in our Pre-Algebra and Algebra forum.

Best Regards,

Mark.