Year: 2013

Acceleration

The following two questions came from a middle-school math textbook. The first is reasonable, while the second is a classic example of an author being overly cute when writing a homework problem.

  1.  A car slams on its brakes, coming to a complete stop in 4 seconds.  The car was traveling north at 60mph. Calculate the acceleration.
  2. A rocket blasts off. At 10 seconds after blast off, it is at 10,000 feet, traveling at 3600mph.  Assuming the direction is up, calculate the acceleration.

For the first question, we’ll assume constant deceleration (after all, this comes from a middle-school textbook). First, let’s convert from miles per hour to feet per second:

60 ~ \frac{\hbox{mile}}{\hbox{hour}} = 60 ~ \frac{\hbox{mile}}{\hbox{hour}} \times \frac{1 ~ \hbox{hour}}{3600 ~ \hbox{second}} \times \frac{5280 ~ \hbox{feet}}{1 ~ \hbox{mile}} = 88~ \frac{\hbox{feet}}{\hbox{second}}

The deceleration is therefore equal to the change in velocity over time, or

\frac{-88 ~ \hbox{feet/second}}{4 ~ \hbox{second}} = -22 ~\hbox{ft/s}^2

Now notice the word north in the statement of the first question. This bit of information is irrelevant to the problem. I presume that the writer of the problem wants students to practice picking out the important information of a problem from the unimportant… again, a good skill for students to acquire.

green line

Let’s now turn to the second question. At first blush, this also has irrelevant information…  it is at 10,000 feet. So I presume that the author wants students to solve this in exactly the same way:

3600 ~ \frac{\hbox{mile}}{\hbox{hour}} = 3600 ~ \frac{\hbox{mile}}{\hbox{hour}} \times \frac{1 ~ \hbox{hour}}{3600 ~ \hbox{second}} \times \frac{5280 ~ \hbox{feet}}{1 ~ \hbox{mile}} = 5280 ~ \frac{\hbox{feet}}{\hbox{second}}

for an acceleration of

\frac{5280 ~ \hbox{feet/second}}{10 ~ \hbox{second}} = 528 ~\hbox{ft/s}^2

The major flaw with this question is that the acceleration of the rocket completely determines the distance that the rocket travels. While middle-school students would not be expected to know this, we can use calculus to determine the distance. Since the initial position and velocity are zero, we obtain

x''(t) = 528

x'(t) = \int 528 \, dt = 528t + C

x'(0) = 528(0) + C

0 = C

\therefore x'(t) = 528t + 0 = 528t

x(t) = \int 528t \, dt = 264t^2 + C

x(0) = 264(0)^2 + C

\therefore x(t) = 264t^2 + 0 = 264t^2

Therefore, the rocket travels a distance of 264 ~ \hbox{feet/second}^2 \times (10 ~ \hbox{second})^2 = 26400 ~ \hbox{feet}. In other words, not 10,000 feet.

As a mathematician, this is the kind of error that drives me crazy, as I would presume that the author of this textbook should know that he/she just can’t make up a distance in the effort of making a problem more interesting to students.

Analog clocks

What time is it?
analog clock
From the perspective of an elementary student — even a good student who generally can read the hands of a clock correctly — the answer probably is 2:50. After all, the hour hand is pointing much closer to the 2 than to the 1.
To address this misconception, perhaps the best suggestion that I’ve heard is using the analogy of waiting for your birthday. A child doesn’t turn 7 until his (or her) birthday comes. On the day before his birthday, he still has to say that he’s 6. Likewise, if the minute hand is just a few minutes before 7:00 and the hour hand is not all the way at seven, we still have to say that the time is six-something.