Question: As much as I agree with your work with keeping it with the real world, all we seem to do in secondary school is things that arent anything to do with the real world, so do you think that you should teach some of that as well?

Mark Hill answered on 16 Mar 2011:

Hi andyisadofinhouspotato,

Bit of an interesting login name you have.

I think that it is vitally important. Take maths for an example. If you have an equation with apparently random, meaningless elements, such as x, y, and z, I find it more difficult to grasp. Use an equation with u, v, a, s and other elements and suddenly there is meaning, of initial velocity, final velocity, acceleration rate (which is co-efficient of friction x gravity) and displacement (distance in a straight line). This gives me one of my most basic equation applications, calculating the speed of a vehicle from a skid mark. Now it all makes sense. Science and maths in the real world.

Please excuse the length of he rest of this answer, but I have given you, in easy terms, an answer to an earlier question, on how to calculate vehicle speed. It is wordy, but simple, and one I use regularly.

From a skid mark? If it skidded to a stop? Or even if it hit something, but that’s a little less straightforward

Let’s start from a common point. The Earth sucks! It really does. Gravity holds us to the ground (most of the time). So any vehicle or object moving should stay on the ground, even after a collision. I expect you are thinking of motorcycles and bicycles, but even then, they, and their riders, ultimately return to the ground.

Now we can use that to help us calculate speed. This looks complicated, but, ask your teacher to write the equation and units on the board and you only have to drop your figures into it. ‘Simples’ (other makes of furry creatures are also available).
In a simple skid to stop of a car we must remeber that the car grips the road because of gravity (g) and the grippiness of the tyres. Gravity is an acceleration rate and is generally about 9.81 metres per second per second (m/s/s). That is an acceleration rate and is important, but put it to one side at the moment.
Let us call the final velocity ‘v’ (speed is almost the same). Then we’ll call the velocity at the start of the skid mark ‘u’. As the car is skidding and losing speed, that is called deceleration (or more accurately, negative acceleration), which is ‘a’.

But to find the velocity of the skidding vehicle I need to know how much grippiness was in its tyres. Broadly all car and vehicle tyres on the same surface, or section of road, have the same amount of grippiness available, or in scientific words, have the same co-efficient of friction. They vary slightly with tyre makes, but that isn’t too important here. However, we’ll call it grippiness here. It doesn’t have any units, but is a ratio, between the tyre and the road surface.
At a collision scene I will usually do a skid test in my police vehicle, with a decelerometer device, which measures how much I am slowing down by. I drive along the road, at about 30 mph, over that collision skid marks… …and skid to a stop. After the tyre smoke has cleared the decelerometer gives me a reading – the grippiness factor – which is called ‘mu’. It is a Greek letter, a bit like a back-to-front ‘y’.

Finally the length of the skid mark, called displacement, is measured, in metres. In equations displacement is always ‘s’.

So, using one of Newton’s equations of motion: u = the square root of (v squared + 2 x mu x g x s).
But our vehicle skidded to a stop, so ‘v’ (final velocity) is zero, and if it equals zero, then we can remove it from the equation, for simplicity.

Now, if you have a co-efficient of friction (grippiness factor) on a typical dry road of, say, 0.7 (remember – no units), gravity is 9.81 m/s/s, and the displacement for the skid mark, say 14.0 metres in our question, then put them into your equation, multiply by 2, then ‘square root’ the answer and you will find that our car was travelling at a velocity of 13.87 metres per second (m/s), which isn’t much use if we want to know whether he was speeding over the 30 mph limit at our scene, at the time of the collision.

A conversion factor, or ‘constant’ can help. We know that if we divide (/) m/s by 0.447, then we can convert it into mph. So, 13.866/0.447 = 31.02 mph at the start of the skid.
Naughty boy.
Buuuuut, that was after he had started braking and wheels don’t immediately start to skid; the wheel start to turn more and more slowly, until they lock-up. Skid marks on tarmac (asphalt) are the result of heat melting the tar, which makes a black line. Very little of the mark is rubber. So, the tyre then needs to get hot enough to melt the tar. Looking at it all, although we can’t measure that part of the event, he was going even faster than the 31 mph that we have calculated.

It all sounds complicated, but it is the really simple way that I work out vehicle speed most days, at crashes. So using ‘boring’ maths, I can sometimes tell what happened in a fatal collision, from the victim’s point of view, when they can’t. Maths in the real world.

I hope that my answer – yes I agree with you – helps. That is why I lecture in schools and colleges. I deliver a couple of hours of what I do, forensics, physics and maths in the ‘real world’, then skid my police Galaxy across the playground, place a shoe on the ground behind it and a red cross in front, to show where a pedestrian supposedly came to rest, and the students then come out, photograph it, measure the marks and look at the evidence.

Then comes the homework! A collision reconstruction, with the maths and physics, to work out if the car was speeding (in the 30 mph limit) at the time it hit the pedestrian. It seems to be good fun for all, as it shows what I do and why.

Good luck.

Mark