Equations of Motion

The 4 equations of motion deal with an object which is travelling with constant acceleration (which can be 0 and therefore constant speed).

The equations are as follows:

(1) v = u + at
(2) s= t(v+u)/2
(3) v²=u²+2as
(4) s = ut + at²/2

Where
a = acceleration – this must be constant for equations to hold
u = initial velocity, ie at the start of the journey
v = final velocity, ie at the end of the journey
s = displacement which is a vector quantity for the distance of the object from its starting point
t = time taken for journey
Any of these values can be found using the equations if at least 3 of the other values are known.

Equation (1) comes from the definition of acceleration as acceleration is the rate of change of velocity and therefore for constant acceleration
a = (v-u)/t which you can rearrange to make
v = u + at

Equation 2 can be found by either considering a distance time graph or using the average speed. (v+u)/2 gives the average speed during the journey as a is constant and buy multiplying this by t we find the displacement. Or from the graph we can find it as the area under the graph is a trapezium of height t and sides u and v so using the area of a trapezium formulae we find 2.

Equation (4) can be formed by substituting 1 into 2 so
s = t(u+at+u)/2
s = t(2u + at)/2
s= 2ut/2 + at²/2
s = ut + at²/2

Equations 3 can also be found using 1 and 2 by rearranging 2 to get an expression for t we find
t=2s/(u+v)

If we substitute this into v = u+at we get
v=u+2as/(v+u)
And by bringing up the v+u we find
v²+vu = u²+vu + 2as
And because we have vu on both sides we can cancel these to find 3

v²=u²+2as

Transformations of Graphs

 

This looks at how a given graph will change when the the function is changed slightly eg how the graph y=x will change when it becomes y = 2x.

y = af(x)

The graph f(x) will "steeper" as the y value of each point is multiplied by a. It will appear like a "stretched" version of the graph y=f(x) 

 

 

y =f(x) + a

The graph f(x) will move up by the amount a as a is added to each y value. This means that the points of intersection of the graph and the y axis will increase by the amount a. The intersection of the graph and the x-axis will depend upon the function of the graph.

 

y = f(ax)

This will make the graph appear "narrower" beacuse y is taking the value of f(x) a time across. So if the value of f(x) is 3 when x =12 then the value of f(4x) = 3 when x=3 as 12/4=3.

y=f(x+a)

This will shift the graph to the left by a. This is because the value of f(x) at x+a is displayed at the point x so effectively the graph occurs a earlier and therefore shifts to the left.

Trigonometry Identities

There a number of “identities” in trigonometry that can be found from the basic ideas of sin, cos and tan as explained in my earlier post. These identities can help in solving equations involving trig functions, especially when there are 2 or more different functions as the often allow you to write the equation in terms of one function, eg sin, that you can then solve.

One of the identities is:

sin² + cos² = 1.

To prove this consider a right angled triangle with side a,b and c as shown below

From this we can use Pythagoras theorem to say:

a²+b²=c²

now we know

sin t = b/c so b = csin t

cos t = a/c so a = ccos t

substituting these values in the above equation we get

(csint)² +(ccost)² = c²

canceling the c² we get

sint² + cost² = 1

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There are trig functions that are equal to 1 over sin, cos and tan called cosec = 1/sin, sec = 1/cos and cot = 1/tan. These can be remembered using the third letter rule as the third letter of each of these corresponds to the the function it is one over.

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Using these a cos² + sin² = 1 we can calculate other identities

tan²t + 1 = sec^2t

We can obtain this by dividing through by cos² as we know sin/cos = tan, cos/cos = 1 and 1/cos = sec.

Other similar identities can be obtained for cosec and cot.

Tan=Sin/Cos

this site is now at www.breakingwave.co.nr

This is often useful when solving trig equations so i thought i’d include it

basically:

sin = opp/hyp

and

cos=adj/hyp

so

sin/cos = ^(opp/hyp)/_(adj/hyp)

so if we cancel the hyp’s we get

sin/cos = opp/adj

and since tan = opp/adj

tan = sin/cos

Sine and Cosine Rules

This is the basics of the sine cos and tan graphs and how sine and cos relate to give you tan. It also shows how to differentiate sin and cos.

The output or range of both sine and cos is from -1 to 1 when given any angle. They can be shown on a graph where y = sin(x) and y = cos(x). In these graphs all the angles go along the x axis and you can see a wave type shape is formed

Sine Graph

Cosine Graph

As you can see both the sin and cos graphs move periodically between -1 and 1 as the angles change, this pattern continues indefinitely because once you pass 360 degrees or 2 pi radians you will return back to the beginning. If you try to perform sin^-1 of a value out side the range -1 to 1 you will get an error.

Differentiate Sin and Cos

also notice that the gradient of the sin graph is the value of the cos graph for the same angle and that the gradient of the cos graph is the -value of the sin graph for that angle. This means that we can differentiate the sin and cos graphs:

if f(x) = sin(x) then f ‘ (x)=cos(x)

and

if f(x) = cos(x) then f ‘ (x) = -sin(x)

however if we use ax instead of x we must differentiate it by bringing the a out, when its just x this doesn’t matter as the differential of x is 1.

ie)

let y = sin(f(x))

now let u = f(x)

du/dx = f ‘ (x)

also

y=sin(u) as u = f(x)

dy/du = cos(u)

from the chain rule

dy/dx = du/dx * dy/du

therefore

if y = sin(f(x))

dy/dx = f ‘ (x)cos(f(x))

and similarly for cos

if y = cos(f(x))

dy/dx = -f ‘ (x)sin(f(x))

Area and Circumference of a Circle: pi

This is a basic guide to using pi to find the area and circumference of a circle using pi. And also explores why pi makes our formulea work.

area =πr²

circumference = 2πr or πd

where r = radius and d=diameter

Area

First lets look at the area of a circle, given by area =πr². This is simple enough to use, we multiply the radius by itself and then by pi.

Does this make sense?

Well r squared is at least going to be an area but it might be a bit small so we multipy by pi. However this doesnt explain much untill we consider what pi is, the easiest way i find to do this is as follows

If we imagine a square that the circle fits inside perfectly(so it touches all four sides like the one above) r squared would give us one quadrant, so the area of that square is 4 x r² . Of course the circle’s area is a bit smaller so we need to find the ratio between the areas of the square and circle. If we then times this value by four we have a magic constant to multiply r squared by to find the area of a circle (we times by four because we need the area of 4 quadrants and r squared gives us one).

Now this magic constant is pi (which makes sense being just over 3, meaning the area of the circle is just over 3/4 of the area of the square).

Circumference

The circumference of a circle is given by 2πr or πd. This seems simple, we just multiply the diameter (2r) by our magic constant pi.

Does this also make sense?

seeing as we only have one r this time so only one length it seems we are just finding a factor to increase the length by to make a different length(the circumference) which makes sense.

Again lets consider the square into which our circle fits perfectly, the perimeter of this square would be 4 time the length of one of the sides.

Now the length of the sides = the diameter so the perimeter is 4d.

Notice again that the value we are trying to find for the square is multiplied by 4, but for a circle were gonna need a ratio thats a bit smaller.

So we need to replace the 4, for a square, with another, smaller, number — it seems pi will do the job.

Conclusion

To me when i consider pi i don’t look at it as a magical fundamental constant, but more a magical fundamental constant multiplied by four, because when I consider how these formula work using pi this is how they seem to work.

So this new constant is really the ratio of

area of square to area of circle

perimeter of square to circumference of circle.

and it = pi/4 = 0.785398….

so if you have a value for a square and you want a similar value for the circle you just need to multiply it by this number and you’ll have your answer

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I welcome comments, improvements or errors in this post. Please leave your comments below or email me at woodford_4@hotmail.co.uk

thanks