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Applying the laws of motions

Recently my A-level mechanics student has had difficulty remembering  the SUVAT equations of motions. I wrote a blog entry on these a while ago.  I’m now trying to think of an easy way to remember them, which a struggle! The best advice is to use them regularly, or just write them out every day.

a = (v-u)/t – is just the definition of Acceleration
s = (u + v)/2 * t  is using the average speed in the old speed * time formula
The other three can be derived from these but in a hurry, better remembered

v2 = u 2 + 2as
s = ut + at2/2
s = vt – at2/2

All of these have different forms where the subject is different, but re-arranging formulas is easier than remembering 20 equations.

There are 5 variables here and five equations..  and each equation has one of the 5 variables missing.  This means we can find the other variables with just 3 pieces of information. Just use the equation with the 3 parts you have and the one you are looking for.

A car travels 20m from a standing start in 10 seconds at constant acceleration. Find this acceleration and the speed after those 10 seconds

We will work out these things separately with two different equations

The acceleration we can find with the 4th equation – the one without v

10 = 100a/2

This is made easier because the standing start means u =0 and we can disregard the first element

10 = 50a  so a = 0.2m/s

We now could use any of the equations to find s as we know everything else, but I do suggest we don’t use a in this calculation,, just in case we didn’t get that right first time – even though I can’;t see any mistake

So we are going to use s = (u + v)/2 * t

20 = v/2 * 10 = 5v

v = 4m/s


Differentiation of the Ln function.

OK  a spot of A-level maths comes to the blog

A few things to remember:

  1. Ln x – The Natural log of x – is the inverse function Footnote 1 of ex
  2. d/dx ex = ex  – That’s what so special about ex

  3. Remember the result from last time that





Now how can we use these three facts to find d/dx ln x?

Let f(x) = y = Ln x . We want f’(x) = dy/dx Footnote 2
Raise e to both sides

ey = eln x  = x  (Using 1 above)

so  x = ey  so dx/dy = ey  (Using 2 above)

so dy/dx = 1/ ey

ey = x so substitute this in the give f’(x) in terms of x and we get 1/x.


d/dx  ln x = 1/x.



1 I use the word function here but caution is required. You must be careful with the domain of Ln x if it’s going to meet the criteria of a Function – It must be defined as x≥0

2. Ok so I’m mixing my calculus formats here! Both have their uses, maybe the subject for another blog entry.

Trig equation of the week….

OK  that title may be a bit misleading  – I’m not proposing a Trig problem every week…  Its also a post more aimed at A-Level students..  I’ll post on more GCSE matters later in the week

I seem to have become a member of a Q&A social networking site called Quora. Now this is very loosely edited, if it is at all, and some of the questions come with strange assumptions (e.g. How many Swedish people really wish they were Americans). The questions on Maths also cover a broad range – One asked how the questioner could find the highest Odd number under 100.

Yesterday though someone posted a question that is right in the frame that my A-level students can have a go at, so I thought I’d reproduce it here…

Rewrite  Sinx + √3Cos x  in the form A sin( x + θ) where θ > 0

OK read no further if you don’t want the answer…

Start by using the expansion of Sin(x + θ) – which by co-incidence I was teaching to a student the day I saw the question.

A sin( x + θ) = A(SinxCosθ + SinθCosx).

Now I’m going to multiply the A into the expression, and re-arrange slightly

ACosθSinX + ASinθCosx.

This is beginning to look like what we need,  but the co-efficients of Sinx and Cosx need matching.  This gives us

ACosθ = 1 (The co-efficient of Sinx)
ASinθ = √3 (The co-efficient of Cosx)

If we divide the second by the first we can eliminate A, for now

Tanθ=  √3..  so  θ = 60 Degrees

Cos60 = 1/2 so we can see from  ACosθ = 1  that A = 2.


And so we have the full solution

Sinx + √3Cos x  in the form 2Sin( x + 60)

Equations of Motion

For today’s post, I’m going to be looking at a subject covered in A-Level.  If that’s not your area of study, you might want to skip this – or read it and see what you learn!

There will be at least one question on an ‘Applied’  or ‘Mechanics’ exam about objects moving at constant acceleration.  There are equations to help with these questions.  One of my students remembers these as the SUVAT equations, as those are the five letters we use.

It pays to remember these for quick question answering, but its also useful and interesting to know where they come from. 

What is Acceleration?

These equations are only for situations where acceleration is constant. Velocity is how quickly distance changes – Acceleration is how quickly velocity changes. An object at rest, or moving at a steady speed are both special cases of constant acceleration – zero acceleration.  And object falling under gravity moves faster as it falls. acceleration is a constant 9.8 ms-2

from this definition comes the first equation;  u is the velocity at the start, v is the velocity at the end (in all these equations). You get the velocity at the end by adding to the velocity at the start the time multiplied by the acceleration

v = u + at

[We always talk about velocity rather than speed, because the direction can change. Velocity is like speed, but includes the direction of travel. Acceleration also has direction. It may sound strange but an object that is slowing down has acceleration  –  in the opposite direction to travel].

Area under a graph

The next equation is easier to find if you draw a graph.

Remember that in a speed Vs time graph, the area under the graph is the distance travelled.
Now we can use the formula for the area of this shape – Base x average height to get

s = t * (u + v)/2

We use s for distance in these equations


The other  three equations

The first two equations are derived by the definitions. There are three more that can be derived from the first two by substitution and algebraic manipulation.  With all five to hand we can choose the right one based on the information provided and required.


Rearrange the first equation  to make t the subject gives t = (v – u) /a

Substitute t for (v-u)/a in the second gives

s = (v+u)(v-u)/2a = (v – u2)/2a

We usually rearrange this so  v is the subject
v = u2 + 2as


The other two we find by making u and then v subject of the first equation and substituting these into the second   – These give

s = ut + at2/2

s = vt – at2/2

That’s our five equations; Look out for the post next week that gives examples on how to put these to use.