# Poorly worded homework problems

A personal pet peeve of mine are grade-school homework problems that are extremely poorly worded, thus leading to unnecessary confusion and bewilderment in students who (sadly) are already confused and bewildered more often than they (or we) would like. Here are two examples that I’ve seen recently.

(1) A worksheet gives the numbers 144 and 300 with the instructions “Find all of the ways to multiply to make each product. First, find the ways with two factors, and then find ways to multiply with more than two factors.”

The second half of the instructions can easily be interpreted by a child to mean “Find all of the ways to write 144 and 300 as a product with more than two factors.” This reading of the question (probably not intended by the author) will take even a gifted child a really, really long time to complete. Furthermore, I’m a professional mathematician, and even I have no idea off the top of my head if there’s an easy formula for the number of ways that a number can be expressed with an arbitrary number of factors greater than 1.

(2) A rocket blasts off. At 10.0 seconds after blast off, it is at 10,000 feet, traveling at 3600 mph. Assuming the direction is up, calculate the acceleration.

I assume that the author was trying to be cute by adding the “it is at 10,000 feet” part of the problem. Or the author wants the student to develop skill at weeding out unnecessary information (like the height) and identifying just the important information (the final velocity and the time) to calculate the quantity of interest.

But it’s aggravating that the information in the problem is not consistent, so there is no solution. In other words, it’s impossible for a rocket to travel with constant acceleration at travel 10000 feet at 3600 mph 10 seconds later.

To begin,

$3600 \displaystyle \frac{\hbox{mile}}{\hbox{hour}} = 3600 \displaystyle \frac{\hbox{mile}}{\hbox{hour}} \times \displaystyle \frac{\hbox{5280 feet}}{\hbox{1 mile}} \times \displaystyle \frac{\hbox{1 hour}}{\hbox{3600 seconds}} = 5280\displaystyle \frac{\hbox{feet}}{\hbox{second}}$.

Therefore, the (presumably constant) acceleration is

$\displaystyle \frac{5280 \hbox{~feet/second}}{10 \hbox{~seconds}} = 528 \hbox{~feet/second}^2$.

However, using calculus, we can compute the height of the rocket by integrating twice:

$v(t) = \int 528 \, dt = 528t + v_0 = 528t$

$y(t) = \int 528t \, dt = 264t^2 + y_0 = 264t^2$

Therefore, the height of the rocket after 10 seconds is $y(10) = 26,400$, not the $10,000$ feet given in the problem.

Next Post