In physics, mathematics is used continuously. As the maths gets more tricky, it seems to lose familiarity and maths classes seem to have nothing to do with the maths in physics.

 Maths classes seem only to use "x, y, z" whereas physics seems to use "s, v, t, a" and other unfamiliar letters which are confusing.

Quadratic equations are usually the first point when physics starts using a "familiar" but otherwise purely "mathematics" type idea and suddenly this hoary old topic becomes unfamiliar.

Basics - A QUADRATIC FUNCTION = anything like

y = 3x2 - 2x + 3 or y = 3 - 2x + 3x2

or y = -3.42q2 + 1.003q

or s = 30.2t - 4.9t2

(NOTICE that the ORDER of writing terms in equations is unimportant, we often forget this, we tend to think that to be a quadratic equation it has to look like ----------.)

In physics, we MODEL real phenomenon. That means we find mathematical equations that "fit" the graphs that measuring nature creates for us. We then use the algebra that "fits" to make new predictions. This is the process of modeling. The algebra we use has EXACTLY the same rules as in maths. That is why you do so much apparently useless stuff in maths ? because it is useful elsewhere, eg physics or economics.


Lets work with the quadratic function

s = 30.2t - 4.9t2

In maths, this is a formula ( a "function" ) which you might be expected to graph.

You would

s = 30.2 x 0 - 4.9 x 0 x 0 = 0 , thus when t = 0, s = 0 so the graph crosses the "s" axis at (0,0) ( OK!  to get nothing, I recognise that nothing times anything is nothing, that is 0 = nothing x something !!!!
               tricky. )

Applying this to the equation, either 0 = 0 x ( 30.2 - 4.9t ) or 0 = t x 0

That is either t = 0 or ( 30.2 - 4.9t ) = 0

If           30.2 - 4.9t = 0  it means that

                       30.2 = 4.9t    ( remember switching addition to the other side changes signs )

           so 30.2/ 4.9 = t ( remember switching multiplication turns to dividing )

giving this answer as t = 6.16.

Two answers for zeroes t = 0 and t = 6.16

The graph is thus an inverted parabola as shown.


In physics, this function models a vertically moving ball which started moving at 30.2 ms-1 upwards and is accelerating down at 9.8 ms-2. "s" is its vertical displacement ( read "height" ) and "t" is the time in the air after starting.

We can use both the graph and the function to calculate useful details.

If we know a value of "t" we can substitute in the function ( or draw on the graph a line ) to find "s". That is - knowing time, we can calculate height.

However, if I know height "s", I can, with a little more difficulty, calculate time, either graphically or through the algebra of quadratic equations.

Why more difficulty?

Look at the example below where I graphically wish to find the times when the height is 20m above the start  ( +20). I get two times, one when going up, the other when coming down.

Now, how do we do this algebraically.

We take the function, s = 30.2t - 4.9t2

And we say, "s = height takes the value, +20" that is

+20 = 30.2t - 4.9t2

that is

+4.9t2 - 30.2t +20 = 0 ( remember rules for moving across equals signs )

What we now have is a QUADRATIC EQUATION.

To solve this little equation - we need either to factorise  ( unlikely to happen easily here)  or to do some fancy work with a rote formula.

Solving +4.9t2 - 30.2t +20 = 0 for t.

Simple versions of these are solved by factorizing, but complicated versions donít easily factorize. We need to use a "formula" approach. In maths one is developed by playing at algebra.

Maths teachers start with the most "general" version of the equation that is possible.

" ax2 + bx +c = 0" ( numbers a, b, c are called coefficients.)

They then show that x will always be calculated by the formula

x = ( -b ± {b2 - 4ac}1/2 ) / 2a                                           remember the indice "1/2" means sq root !

In the above case a = +4.9, b = - 30.2, c = +20 so

t = ( - -30.2 ± {[-30.2]2 - 4x4.9x20}1/2) / 2x4.9

= (30.2 ± {912 - 392}1/2 ) / 9.8 = (30.2 ± 520 1/2) / 9.8 = ( 30.2 ± 22.63) / 9.8

giving two answers which can be seen on the graph above

t = 5.39 seconds and 0.77 seconds.

The 0.77s corresponds to going through +20m on the way up

while the 5.39s corresponds to +20 m on the way down.

 This is clear from the graph above.


Calculate the times for the function  s = 30.2t - 4.9twhen s = +10m and when s = - 10m . See if these agree with the graph.

Sometimes the answers will not make sense meaning that no real answers exist. An example would occur for our  s = 30.2t - 4.9t2 if s = + 60m.    The algebra would give you ridiculous square root answers. Physically it means that the ball does not reach +60m. ( Check both by calculating using the formula and by looking at the graph).

At present in physics you will not meet these cases.

Not all answers have physical meanings. Negative time answers are simply artificial artifacts of using algebra and not looking at the physical situation.

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