Today, I thought I would share my experience using Newtonian mechanics over the past week to teach introductory algebra. While there is no way to avoid drilling the fundamentals (e.g. operator precedence, substitution, etc.), there is absolutely no reason to teach algebra abstractly. It is a FACT that most children learn better by relating new concepts to concrete ideas they already know. Paulina is no exception. She loves science. Hence, it makes sense for me to pick problems and examples from physics and astronomy as a vehicle for learning algebraic concepts.
We are studying linear and simple quadratic equations. To do this, I am using the linear relationship between Celsius and Fahrenheit (e.g. \(F = \frac{9}{5}C +32\)) and the following three equations from Newtonian mechanics:
- \(d = vt\)
- \(v = at\)
- \(d = \frac{1}{2} at^2\)
Let me start with the motivation for the Celcius-to-Fahrenheit conversion. We moved to Amsterdam from Los Angeles last July. While the weather is beautiful where we came from, temperatures are quoted in Fahrenheit and rarely dip below sixty degrees. This means that Paulina has very little idea of what zero Celcius feels like. Since we are learning about the slope-intercept form of a line (e.g. \(y = mx + b\)), I gave her two (Celcius, Fahrenheit) ordered pairs and asked her to derive the equation to convert from Celcius to Fahrenheit. In particular, I gave her \((0,32)\) and \((10, 50)\). The point of this problem was to make her work on a result she could use to make sense of temperature in our hometown. Next, I asked her to compute the Fahrenheit boiling point of water in (For reasons unknown to me, few people in the US know the boiling point of water in Fahrenheit, but everybody knows that it freezes and boils at zero and a hundred degrees Celsius respectively!)
This problem provides a few fringe benefits:
- It is practical and relates to something concrete that my child cares about
- It requires knowing how to find the slope
- It requires knowing how to find the y-intercept
Let me continue now with how we used Newtonian mechanics. Bear in mind that we have been working on and talking about Newtonian concepts for a while. You may have to start at a different point than I illustrate here. I told Paulina that a guy jumped off a plane as fell for two minutes and that his parachute did not open. Then, I asked her to compute how fast he was going when he hit the ground and how high he was when jumped off the plane.
This problem requires three things:
- Knowing which equations to use
- Knowing what data to plug in for what variable
- Interpreting the answer
The most important part of this problem is analyzing the answer to determine if it is sensible (e.g. was the question sensible to begin with). Let me solve the second part. How high was the man before he jumped off? Clearly, one must use \(d = \frac{1}{2} a t^2,\), with \(a = g = 9.8 m/s^2\). This implies that $$d = \frac{1}{2}at^2 = \frac{1}{2} 9.8 \frac{m}{s^2} \cdot \left(120 s\right)^2 = 70,560 m.$$ The key observation here is that \(70,560 m\) is \(70.5 km\). The problem is that skydivers typically jump from 5,000 to 15,000 feet and commercial airliners fly at 35,000 feet. Hence, there is no way for somebody to free fall for two minutes, unless something slowed down the descent (e.g. drag). This is most important message to drive here. It is just as important to know how to solve the problem as it is to think about the answer and what it means.
The key message of this post is that it is highly effective to ground the teaching of mathematics on concrete subjects that your child cares about. Not only will your child learn math better, but he or she will also gain a deeper understanding of the subjects use to motivate the math.





