# Physics, the Car, and the Hill

Warning: this post will contain some basic physics. Read at your own risk 🙂

One of my favourite units in physics was kinematics. I loved the simple formulas and precise answers I was able to obtain for almost any problem. I also liked how I could take a complicated situation and model a much simpler version of it using kinematics. So as I was driving up a hill the other day I had a question pop into my head. Here is a diagram.

Suppose I am driving up to a hill, and I am travelling 60 kilometres per hour. Now I take my foot off the gas and coast up the hill. When I reach the top, I have lost some speed and I am traveling 40 kilometers per hour. This on its own seems like a pretty simple situation. However, here was where my question came in.

Now suppose right after the first hill, there is a second hill, the same height and incline as the first.

My question is: Can I make it up the second hill?

Think about it for a bit. Intuition would say “of course you can.” I was traveling 60 km/h at the bottom of the first hill and 40 km/h at the top. That means I lost 20 km/h by going up the first hill. So I would assume that when I go up the second hill I will lose another 20 km/h and end up going 20 km/h. In fact, intuition would tell me that I could probably make it up 3 hills the same height before I run out of speed and stop.

However, this is not the case. The car will only get 70% of the way up the second hill before stopping (and then sliding back down, poor car). Here is the math. If you want to skip the math, I have included an intuitive explanation in the conclusion for the more linguistic readers.

We need to use 2 formulas and 1 physics law. We will be dealing in terms of energy. The first formula is for kinetic energy. This is the energy an object has when it is moving. It states that the energy of a moving object is:

$KE = \frac{1}{2} * m * v^2$

KE: the kinetic energy of the object in joules

m: the mass of the object in kilograms

v: the velocity (or speed) of the object in metres per second

The second formula is for potential energy. This is the energy an object has because of its height above the ground. It states that the energy of an object is:

$PE = m*g*h$

PE: the potential energy of the object in joules

m: the mass of the object in kilograms

g: the speed of gravity (9.8 m/s2)

h: the height of the object above the ground in metres

The physics law we will use is the law of conservation of energy. It states that when you add up all the energy (in our case potential energy and kinetic energy) of a system, you should have the same answer, no matter when you do the adding. In other words, the car should have the same energy at the bottom of the first hill as it does at the top of first hill and at the top of the second hill (if it can even get there).

Ok, let’s get cracking. An average car weighs 1000kg. We will call the bottom of the first hill 0 metres in height. We have to convert our speeds from km/h to m/s. Here is the algebra to find the height of the first hill:

$KE1 + PE1 = KE2 + PE2$

$\frac{1}{2}*1000*17^2+0 = \frac{1}{2}*1000*11^2+1000*9.8*h$

$h=8.6m$

Now to find the velocity at the end of the second hill

$KE2 + PE2 = KE3 + PE3$

$\frac{1}{2}*1000*11^2+ 1000*9.8*8.6 = \frac{1}{2}*1000*v^2+1000*9.8*17.2$

$v= \sqrt{-47.6}$

We get the square root of a negative number. Since this gives us a complex velocity (which is impossible in classical kinematics) we know it must be impossible for the car to make it up the second hill. Simple algebra (left as an exercise for the reader) tells us that the car can only get up to a maximum height of 14.7 m or about 70% the way up the second hill.

In conclusion, the reason the car does not make it to the top of the second hill is that speed and energy are not related in a linear way. If my car, going 60 km/h, loses speed and ends up going 40 km/h, this is not the same as my car, going 40 km/h losing speed and ending up going 20 km/h. We say that speed and energy have a “power relation.” It’s that pesky 2 in the formula that messes with our intuition.

$Energy \approx speed^2$