## WebWork Section 13.2

The due date for WebWork section 13.2 has been changed to Wednesday night. (Normally, WebWork assignments posted on Friday are due on Monday night.) This change just reflects that we didn’t have class on Monday.

### Problem 3

I have given everyone credit on problem 3. My version of the problem (we all have slightly different versions from WebWork) says the following: A plane is heading due south and climbing at a rate of 110 km/hr. If its airspeed is 510 km/hr and there is a wind blowing 100 km/hr to the northwest, what is the ground speed of the plane?

First, I’ll give you my solution and then explain why WebWork asked the question incorrectly. I should point out before I begin that many students were able to answer this correctly. But, many also had questions due to the wording.

### Solution

We’ll view the xyz-coordinates as usual and view east as the positive i-coordinate, south as the negative j-coordinate, and so on. If the plane is heading directly south, then its overall ground velocity vector has an i-component of zero. Let’s break this in to three parts. (1) The northwest blowing wind has a velocity (as viewed from the ground) of

$\vec{A}=-\frac{100}{\sqrt{2}}\vec{i}+\frac{100}{\sqrt{2}}\vec{j}\ \text{km/hr.}$

(2) The plane has a component velocity (air-speed, as viewed in the air) of

$\vec{B}=a\vec{i}+(-b)\vec{j}+100\vec{k}\ \text{km/hr}$

which equals out to 510 km/hr. (3) Overall, these velocity vectors (and the fact that the plane is heading south) imply that

$\vec{A}+\vec{B}=0\vec{i}+\left(\frac{100}{\sqrt{2}}-b\right)\vec{j}+100\vec{k}\ \text{km/hr}.$

Solving for $a$ and $b$, and then using only the i and j-components of the right side of the last equation will tell you the ground speed.