# High School: Number and Quantity

### Vector and Matrix Quantities HSN-VM.A.3

3. Solve problems involving velocity and other quantities that can be represented by vectors.

Students should understand that verbal problems aren't going to kill them, make them vomit, or cause them any permanent brain damage. Of course, they can always consult their physicians about that, too.

Verbal problems are the real "guts" of most math courses. After all, how often will they have to solve the equation x2 + 8x – 3 = 18? Not too often, right? (With the exception of becoming a math teacher. Well, you know all about that, don't you.)

But verbal problems are a different kettle of fish. They give students the opportunity to put all those equations and logical thought to practical use. Students should be able to take the concepts of vectors and apply them to velocities in real-life scenarios.

Let's assume you own a boat (and hopefully that's a real-life scenario). You're on a river that has a current of 4 mph and your speedometer says you're going 15 mph. If you're going downstream, how fast are you actually going?

Going downstream means that the current is actually pushing you where you want to go, increasing your speed. The magnitude of your boat's vector is 15. The magnitude of the current is 4. The combined magnitude, the sum of those two vectors going in the same exact direction, is 19. That means you're actually going 19 mph.

On the return trip, you're going upstream under the same conditions. Now how fast are you actually going?

Well, you knew it was too good to be true. That same current that helped you make time on the trip downriver is now pushing against the front of your boat. The boat's vector and the current's vector are pointing in opposite directions.

Now your 15 mph magnitude is being added to a -4 magnitude, for a resultant of only 11 mph. Even though your boat's engine is chugging along at 15 mph, you're only achieving 11 mph. The current is slowing you down.

What about more complicated problems where vectors aren't opposite one another? Students should be able to use a right triangle or a coordinate plane to solve slightly more complicated vector word problems, such as finding the velocity of a plane flying due south at 100 mph with a side wind of 20 mph, going due east. (The answer, by the way, is , or approximately 102 mph.)

#### Drills

1. Ted is on a sled, going downhill at 20 mph. He tosses a snowball down the hill, throwing it with a velocity of 5 mph. Find the speed of the snowball.

25 mph

Since both vectors share the same direction, their combined magnitude is just the sum of the individual magnitudes. In other words, because Ted is already going the same direction as the snowball he's throwing, the two speeds add up. Since we know from first grade that 20 + 5 = 25, (A) is our answer.

2. Ted's brother Ned, on a sled ahead of Ted, did not appreciate the snowball. He lobs one back at Ted. Ned's sled is traveling at 23 mph, and he throws it at a velocity of 9 mph. Find the velocity of the snowball as it flies uphill through the air.

14 mph

This time, the vectors are going in opposite directions. The sled is going downhill while the snowball is being tossed uphill. That means we subtract the magnitudes, for a velocity of 14 mph uphill.

3. Alexandra flies her plane due east and the speedometer says 150 mph. If there are winds traveling west at 25 mph, how fast is Alexandra's plane actually flying?

125 mph

Alexandra is flying her plane against the direction of the wind. That means her speed is being reduced by the speed of the wind. In other words, we subtract 150 – 25 = 125 mph. So Alexandra's plane is only going 125 mph.

4. If Alexandra slows down until she's flying eastward at 100 mph exactly, what would her speedometer say? Assume the winds are still traveling west at 25 mph.

125 mph

Since Alexandra's true speed is the combination of her plane's velocity and the velocity of the wind. Since the wind is traveling in the opposite direction at 25 mph, Alexandra has to push forward an extra 25 mph to counteract the wind's vector. That means her speedometer will read 125 mph when she travels at 100 mph.

5. John is driving his boat due north, with a velocity of 24 mph (or so it says on his speedometer). There's a current going due east, with a rate of 7 mph. Find John's actual velocity.

25 mph

North and east aren't opposite each other; they're perpendicular. That means we can use a right triangle to help us solve the problem. The boat's speed is the height of the triangle and the current is the vector along the base. We can use Pythagorean theorem (or, for a shortcut, we can realize that we have a 7-24-25 Pythagorean Triple). The resultant vector is 25 mph.

6. A wind has blown Kelly's sailboat off course. She had hoped to sail due north at a rate of 9 mph. A westerly wind blew her off course, however, with a resultant velocity of 15 mph. What was the wind speed?

12 mph

This is another example of a problem that needs a right triangle. Kelly wanted to go due north at 9 mph, which translates to a vertical vector of length 9. The westerly wind diverted her, making her travel at a slanted vector at 15 mph. We are given the height and the hypotenuse of the right triangle, and need to find the horizontal base. Fortunately, it's another Pythagorean Triple and our answer is (C).

7. In problems where the vectors are perpendicular, why is it important to know the direction of the current?

You need right triangles to do Pythagorean theorem

In triangles that are not right triangles, you would probably want to use trigonometry—the Law of Sines or the Law of Cosines, for example. While it's possible to create more complicated problems using these trigonometric laws, the Pythagorean theorem is easy and simple and it won't get our knickers in a twist.

8. Vlad and Markus are playing a nice, safe game of paintball. Vlad shoots a paintball at Markus at 60 mph due south. If the wind is blowing in the exact opposite direction (due north) at 2 mph, what speed does the paintball travel?

58 mph

If the paintball is released from the barrel at 60 mph, but the wind is blowing in the opposite direction, the vectors for the velocity of the paintball and that of the wind are opposite one another. That means we have to reduce the speed of 60 mph by 2 mph, and of we do so, we end up with (B).

9. If Markus shoots a paintball back due north at 55 mph and the 2 mph wind is blowing with the paintball, how fast is the paintball going?

57 mph

Since the paintball is traveling along with the wind, the two vectors with the same direction compound. That's fancy talk for we add 'em up. We're well aware that 55 + 2 = 57, right? So that's our answer.

10. Vlad and Markus don't stay in one place during paintball, though. That's a surefire way to get hit over and over, and they each want to win. Vlad then fires another shot at 59 mph due west at Markus. If the wind has increased to 19 mph (that was fast!) blowing due north, how fast is the paintball really traveling?