A common problem for teaching Physics when you are not a Physics teacher is that you make many mistakes with complex ideas which appear simple on the surface.
I drop a ball to the earth and want to work out the distance fallen in a certain time. I then do a calculation of..
distance x speed = time
Oh dear but when I do the experiment it does not work like that but it works like this…
[latex size=”3″]s =\frac{1}{2}at^{2}[/latex]
Now we can reason this out. If I allow a ball to fall to earth it must accelerate due to the field of gravity around the earth 9.81N/kg. Or 9.81 m/s/s.
But where does the formulae come from that we general use?
[latex size=”3″]s = ut+\frac{1}{2}at^{2}[/latex]
This post is simply to give some advice about this formulae. To start with define everything we use…
- s = the distance between initial and final positions (displacement) (sometimes denoted as x)
- u = the initial velocity (speed in a given direction)
- v = the final velocity
- a = the constant acceleration
- t = the time taken to move from the initial state to the final state
So the question is what is it all about.
Well we need to think way back to the idea of a simple idea of how to work out the distance travelled by a runner in a race.
Think of an athlete travelling 100m in 10s at a constant speed. His velocity or speed is…
[latex size=”3″] \frac{d}{t}= speed[/latex] OR [latex size=”3″] \frac{s}{t}= v[/latex]
Now this works fine if we are travelling at a constant speed but hey as you know this is not always the case and sometimes an object has a constant acceleration or deceleration. If you think about a graph of a person who got faster and faster then we would have a slope or triangular area on a speed-time or velocity-time graph. Now the area under the graph would be the distance travelled on the journey. But if our speed changed s=vt (for constant speed) becomes….
[latex size=”3″] s=\frac{vt}{2}[/latex]
Now you have an expression for the average speed but only from a standing start. Imagine now the same velocity time graph but this time the runner was already travelling at a velocity at the start of timing. Our area would become a triangle and rectangle…
[latex size=”3″] s=\frac{(v-u)t}{2}+ut[/latex]
Simplifies to
[latex size=”3″] s=\frac{(v+u)t}{2}[/latex] – Eq 2
This is the formulae for average speed that takes care of all situations even when u or v is 0. Now then think graphically again if we are travelling at an initial velocity and then accelerate we are back to triangle and square again when looking at a vt graph. Hence…
[latex size=”3″] v=u+at[/latex] – Eq 1
Now we can use this in a rearranged form..
[latex size=”3″] t=\frac{(v-u)}{a}[/latex]
Substitute into Eq 1 in new form into 1 to remove t so we now have an expression…
[latex size=”3″] s=\frac{(v+u)}{2}*\frac{(v-u)}{a} [/latex]
Which simplifies to..
[latex size=”3″] s=\frac{(v^{2}-u^{2})}{2a} [/latex] – Equ 4
Now we have this we can work also work backwards to get Eq 3 …
[latex size=”3″] v=u+at[/latex] – Eq 1
And *t gives..
[latex size=”3″]vt = ut+at^{2}[/latex]
Divide both sides by two…
[latex size=”3″]\frac{vt}{2} = \frac{ut}{2} +\frac{ at^{2}}{2} [/latex]
Add ut to both sides, multiply by 2 and tidy up..
[latex size=”3″]s = ut+\frac{ at^{2}}{2} [/latex] – Eq 3
So we now have the four equations of motion for an object which is travelling at a constant velocity or accelerating at a constant rate. As shown you can reason them all out with very simple ideas from first principals. You can also call them whatever number you want as some people label them differently.
So hopfully we have now understood where the formulae comes from and realise that if we drop a ball we must apply the formulae to work out the distance fallen in a more complex way.
Then it gets really interesting when you think about a tanker travelling at a constant velocity which then slows down. How far does it travel as it slows…
[latex size=”3″]s = ut-\frac{ at^{2}}{2} [/latex]
Of course, it will be ut for the whole time and then you take away a little bit of “s” from the other term as you slow.
