## Learning Math: Patterns, Functions, and Algebra

# Linear Functions and Slope Homework

## Session 5, Homework

**Problem H1**

In the Achilles and the tortoise problems in Part C, Achilles runs at a constant rate of 9 miles per hour, and the tortoise moves at 1 mile per hour. Suppose that the speeds of Achilles and the tortoise are unchanged but Achilles catches up to the tortoise in 1 1/2 hours. How much of a head start did the tortoise get?

*Tip: Using a spreadsheet can help solve this problem. *

**Problem H2**

The tortoise has taken some “turtle speedup potion” and can now walk at 2 miles per hour. If Achilles still runs at 9 miles per hour and catches up to the tortoise in 3 hours, how much of a head start did the tortoise get?

**Problem H3**

Here’s a trick that master carpenter Norm Abram might use when building supports for roofs. He knows he’ll need evenly spaced supports along the roof. He carefully measures what length he needs for the 1st one, and finds that it’s 12 feet. Then he measures what he’ll need for the 2nd, and finds it is 9 feet. He calls to his assistant: “Don’t measure the others, just make them 6 and 3 feet long!” Why does Norm’s trick work?

**Problem H4**

You’ve worked with undoing functions. Take a moment to think about undoing a linear function. If given the formula d = 3t + 2 for distance traveled in terms of time, what would you do to express time in terms of distance? When undoing a linear function, will the result always be a new function? If so, will the new function always be a linear function?

### Solutions

**Problem H1**

Earlier we found that Achilles ran 13 1/2 miles in 1 1/2 hours. Therefore, we have to find a way to get the tortoise to the 13 1/2-mile mark after 1 1/2 hours. The tortoise walks at 1 mile per hour, so it can walk 1 1/2 miles in that time. The remaining distance must be its head start: 13 1/2 – 1 1/2 = 12 miles.

**Problem H2**

Use the same logic as in Problem H1. Achilles runs 27 miles in the 3 hours, therefore the tortoise needs a head start that will get it to the 27-mile mark after 3 hours. Because it walks at 2 miles per hour, it can walk 6 miles in 3 hours, so the head start is 27 – 6 = 21 miles. An algebraic equation could also be used for this problem.

**Problem H3**

It works because the roof is a straight line, and therefore it has a constant rate of change. By carefully measuring the 1st and 2nd support, the carpenter has calculated a rate of change: (change in height of support) / (distance between supports). Since this rate is constant, and the distance between supports stays the same, the change in the support’s height must also be constant. This is identical to predicting the next number in the output of a linear function; in this case, the output drops by 3 for every new support.

**Problem H4**

It can be done by solving the algebraic equation for the other variable, using the technique of “undoing” that was first used in Session 3. For the equation d = 3t + 2, start by subtracting 2 from each side to produce d – 2 = 3t. Then divide both sides by 3, so that the equation is (d – 2) / 3 = t. If a linear function can be undone, the result will always be a new, linear function. The only linear functions which cannot be undone are constant functions like y = -7. See Session 5, Problem D7 and Session 3, Problems E9-E12.