## XUVAT – it’s not just a good idea, it’s a rule!

So you have learnt your 5 equations of motions, but do you really understand them? Most of you are probably nodding your heads… and those of you who aren’t should be working towards making sure you can be nodding! (please note that sometimes these are called the SUVAT or DUVAT equations as well)

You might be surprised, then, to discover that what you have learnt is only the very basics of the “Kinematics Equations”. There is much more for you to learn as you go through physics and Mathematical Methods. Let’s review some of the basics.

First and most important these equations are the “Constant (or uniform) Acceleration Equations”. This means that they only work if the acceleration of the body is constant. This means that you cannot use them if the object is increasing or decreasing it’s acceleration, i.e. it has non-zero jerk.

Secondly, you have to be careful to use the scalar version when dealing with scalars, and vector version when working with vectors – but remember that adding and subtracting vectors has special rules.

Finally, remember to use your DATA BLOCKS. Data blocks are essential to make sure you don’t make simple mistakes – and they help you earn marks in assessment (tests, exams and assignments). Speaking of which, here is the test for redoing – make sure you show a data block every time you have to use one of the XUVAT equations.

We know where some of the equations come from, and I’ve shown you one of the more complex ones. Here is a video that explains the rest:

So where to from here? You need to understand the relationship between each of the kinematics graphs (x against t, v against t and a against t), and also how they relate to the XUVAT equations. In order to do this, I want you to work completely through these two sites (this is not a work requirement, but if you are thinking that doesn’t mean you have to do it, you are absolutely wrong – and I will be able to tell.) (Yes, that is a threat… or a hint) (Hey what was that? Behind you! Something captured my attention!)

And here are two links to java applets that will allow you to investtigate some of the interactions between speed-time and distance-time graphs (1,2)

See you in class!

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### 15 Comments on “XUVAT – it’s not just a good idea, it’s a rule!”

1. David G Says:

Is it possible to save the test in a .PDF Format and on Google Docs for easier access and downloading? I’m having trouble opening the link and saving/printing the test out.

2. David G Says:

Can i get the link to the tracker program? If it isn’t hidden somewhere on this page…

3. Rares Luca Says:

Sir we have one Sac here but we need a copy of the other one. Can we get a link or something. I am a bit confused. I’m not sure but apparently all work requirements are due tomorrow. Also sir I must admit… We miss u… That was hard to type…. ðŸ˜€

Anyway if we can get a link on the other sac please also a update on the latest news regarding physics work and work requirements… ðŸ˜€ Thanks and peace out .

• CyberChalky Says:

Hi Rares,

Sorry this is so late; please spread the word to as many as possible: The Astrophysics test is here.
I’m still very sick – I’m hoping to be back on monday, but now the doctors are taking chest x-rays and talking about sending me to specialists…

4. Paktia Says:

Also, i e-mailed my prac to you, hope you received it.

5. dylaniscool Says:

hey sir, i was away today so i couldnt show you the work requirements, do you want me to scan them and send them to you? The only one i havent done is the vector prac because i was away and have tried to get the info of others but no one seems to have it.

6. darrren10 Says:

Hey Mr G,
A girl with mass 50kg running at 5m/s jumps onto a 4kg skateboard travelling in the same direction at 1.0m/s. What is their new common velocity.

^ How would we begin to work that one out?

• darrren10 Says:

Actually I think I just had an epiphany and worked this out. The answer I got was 4.7m/s.

The way I did it:

P=mv

m(girl) X v(girl) + m(skateboard) X v(skateboard) = Total Momentum

= 50X5+4X1
= 250 + 4
= 254 kgm/s

Since momentum is conserved then after she jumps onto the skateboard the momentum should be the same.

54 X v(combined) = 254
v(combined) = 254/54
=4.70m/s

Is that correct?

• CyberChalky Says:

Exactly. You can push yourself a bit further if you do the “ice skater” type problems, where two individuals are at first skating together (houlding on to each other), and then push off from each other.

7. […] Motion: XUVAT equations, Centre of mass, Energy transformations, Basic […]