Help with this equation please (1 Viewer)

[golfpro]

I was asked in my work to show how this equation works or whatever;

r = ut + 1/2at²

for r = total displacement.

now i understand that the total dispacement is equal to the initial velocity = u x t = time , which gives us our first displacement and the second displacement is given by the acceleration x the time.

but iundertand this except for why the half the acceleration and then square the time. that completely confuses me.

can someone explain why they half the acceleration and square the time for the second half of the equation for the total displaccement.

 

[golfpro]

also another which says show how this equation works when the initial velocity = 0


t = √2r
------
a

now the square root is supposed to encapsulate all that decimal bit but im crap with the computer.

 

[watatank]

You mean derive it? You don't have to be able to derive that!
s = ut + 0.5at^2

s = displacement
u = initial velocity
a = acceleration
t = time

There's a way to explain this...but you need to know calculus....which you never use in HSC physics.

 

[watatank]

also another which says show how this equation works when the initial velocity = 0


t = √2r
------
a

now the square root is supposed to encapsulate all that decimal bit but im crap with the computer.

Well if u = 0,

s = 0.5at^2

simply rearranging the equation.

2s = at^2

2s/a = t^2

t = sqrt(2s/a)

 

[falcon07]

r = ut + 0.5at^2

If you draw a graph of velocity vs time, the area underneath the line will be equal to the displacement. If you are dealing with constant acceleration, acceleration will be the gradient of the graph. Dividing the area under the graph into a rectangle and a triangle, you get a rectangle of height u (initial velocity) and width t, so a displacement of ut. The height of the triangle is given by at (gradient multiplied by x axis length), so the area of the triangle will be 1/2 base x height, or 0.5at^2. Adding these 2 areas gives you the formula.

 

[alcalder]

But on the calculus, if you know how to use it, you certainly can use it in Physics. It is not wrong to do so.

Knowing that...

In a vertical direction acceleration is always:

a_y = -g (due to gravity and downwards)

Integrate this to get velocity:

since a_y = dv_y/dt = -g

v_y = -gt +CONSTANT

If we consider an abject that at time t=0 starts from velocity of u, put that into the equation above to find that CONSTANT = u

THUS

v_y = u - gt

BUT v_y = ds_y/dt = u - gt (where s = total displacement)

Integrate again

s_y = ut - 1/2 gt² + CONSTANT

But at time t=0 our body starts from the origin so CONSTANT = 0

THUS

s_y = ut - 1/2 gt²

Now, remember we said that a_y = -g

Then

s_y = ut + 1/2 a_yt²

YOUR EQUATION. That's how it is derived using calculus, in case you wanted to know.

 

[Forbidden.]

So diffrentiating s_y = u_yt + 1/2 a_yt² gives v_y = u_y + a_yt (is it ?) but why just v = u + at on the HSC Physics formulae sheet ?
Because I didn't use v_y = u_y + a_yt and it was critical to getting the right answers too ...

 

[alcalder]

Yes, you can differentiate and get

v = u + at

This is one of the simple formulae just given when doing motion in physics. OR in another guise.

a = (v-u)/t (which is essentially average acceleration = change in velocity over time - oh look: a = dv/dt *snigger*)

The little _y thingies are left out, probably, because this formula holds for all motion in all directions when considering accelerated motion. It only gets specific for the y-direction when you start introducing a=-g.

I guess the formulae they decide to put on the HSC formulae sheet are those mentioned directly in the HSC syllabus. I always maintain, you need to know what those formulae mean and how to use them and all the circumstances where they can be used.

 

[watatank]

To add to alcalder's calculus derivation, if it wasn't obvious you integrate (or differentiate) with respect to 't'