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# M1 maths Vectors Motion in 2 dimensions (AQA) watch

1. The problem I am trying to solve is from AQA Mechanics M1 and is as follows:

An object is moving in a plane. At time t=0, it is at the origin, O, and moving with velocity u. After 2 seconds it is at A where OA = -2i - 4j. After a further 3 seconds it is at B where AB = 10i - 40j. Show that this is consistent with constant acceleration a, Find a and u.

I don't understand what "Show that this is consistent with constant acceleration a"means, Do I have to find the acceleration between O and A and then between A and B and work out if they are the same? I've tried this with the initial velocity of u to find the acceleration for the first step and the velocity at the end of the first step then I tried to use the velocity as the start point for the second step with a different a hoping it would come out as the same value as a in the first step. The equations were enormous and I don't think this was the right approach. However, this question being unlike any example or other exercises in the chapter, I'm at a loss to know where to start.

Any clues?
2. (Original post by maggiehodgson)
Any clues?
If, and only if, the information is consistent with constant acceleration, then suvat will model it.

So, if you can find a,u such that s=ut+1/2 at^2 is true for all points O,A,B, then you've shown it's consistent, and found u,a in the process.

s,u,a are of course vectors here.
3. For the first leg of the journey I get (when t=2 u is the value u and a is the unknown acceleration)/

-2i - 4j = 2u + 2a (a= -i -2j -u)

Then I can find v = -i -j -u and this becomes the starting velocity for the leg 2 of the journey.

12i - 36j = 2(-i -j -u) - (1/2) 9(new acceleration which we are trying to show is the same as in leg 1)

2(14i -34j +2u)/9 = new a

Surely a for leg 1 and a for leg 2 are different?

Can you give me more clues please?
4. (Original post by maggiehodgson)
For the first leg of the journey I get (when t=2 u is the value u and a is the unknown acceleration)/

-2i - 4j = 2u + 2a (a= -i -2j -u)

Then I can find v = -i -j -u and this becomes the starting velocity for the leg 2 of the journey.
Don't think your value for v is correct.

12i - 36j = 2(-i -j -u) - (1/2) 9(new acceleration which we are trying to show is the same as in leg 1)
Whilst the last term is correct there, the other two are in error.
5. Thanks for all this help you're giving. What was I thinking.

if I use r = (u+v)t/2 to find the velocity at the end of leg 1 get
-2i -4j = u+v
v =-2i - 4j -u

Then I use v = u + at to find a for leg 1
-2i -4j -u = u +2a
a= -i -2j -u

I then use the v value as the starting value for leg 2
and use r = ut +.5at^2
Am I right in thinking that r = (10i + 2i) - (40j +4j)?

If so
12i -36j = (-2i -4j -u)3 +0.5a9
4i - 12j = -2i -4j -u + 1.5a
(6i - 8j +u)/1.5 = a

I'm really stuck but if you can put up with me I'd be pleased to hear from you again
6. (Original post by maggiehodgson)

if I use r = (u+v)t/2 to find the velocity at the end of leg 1 get
-2i -4j = u+v
v =-2i - 4j -u

Then I use v = u + at to find a for leg 1
-2i -4j -u = u +2a
a= -i -2j -u
Yep.

I then use the v value as the starting value for leg 2
and use r = ut +.5at^2
Am I right in thinking that r = (10i + 2i) - (40j +4j)?
Nope.
For leg 2 we're going from A to B, and you're told what AB is in the question (unless that's a typo?).
7. Had to go to work so sorry for late reply. Thanks for that. Foolishly, I was mistaken in what the AB vector meant.

So for the second leg of the journey
r = 10i- 40j?
u = -2i -4j -u?

If I now use r = ut + 0.5at^2
I get

10i - 40j = (-2i -4j -u)3 + 0.5a9
10i - 40j = -6i -12j -3u +4.5a
16i - 28j +3u = 4.5a

This does not turn into the a for leg 1.

I'm not cracking this at all am I.
8. (Original post by maggiehodgson)
I'm not cracking this at all am I.
You're doing fine.

This does not turn into the a for leg 1.
You now have two different representations for a.

If this is constant acceleration, those two representation. are equal, and you can solve for u, and a.

You can then construct the s=ut+1/2 at^2 equation and check the position at times 0,2,2+3.
9. I have to admit, after an hour's worth of trying, that I am no wiser.

I am working on the idea than when you say "those two representation. are equal" it means that they are interchangeable. So I've tried putting the other a into s = ut + 0.5at^2 but it has got me nowhere. Then I tried combining the a's to see if that helped but, again, nothing.

I've worked out the velocity at the end of leg 2 and then "assuming constant acceleration" worked out how far it had travelled divided by 5 and multiplied by 2 to see f that was anything like r at the end of leg 1. No it wasn't.

I then looked at the answer in to book to see if I was getting anywhere. I look to be in no-man's land.

Would you mind just starting me off and I'll give it another go.

Thanks

Maggie
10. (Original post by maggiehodgson)
Would you mind just starting me off and I'll give it another go.
From leg 1 you have: a= -i -2j -u

From leg 2 you have : 16i - 28j +3u = 4.5a

You now need to solve these as simultaneous equations. There are only 2 unknowns, a and u.

You will then have the values to plug into s=ut+.... to confirm the 3 points O,A,B.
11. Wow!

Thank goodness for you and thank goodness I've got there. That was so hard/complex for chapter 2 book 1! I think it was anyway.

Thank you so very much for all your help and encouragement. "Statics and Forces" here I come. I hope that if I ever post again that I'm lucky enough to get you watching.
12. (Original post by maggiehodgson)
...
You're welcome.

One small thing. Rather than work with leg1 and then work with leg2 to create your two equations.

I'd work with leg1, and then I'd work with leg1 and leg2 combined into one continuous interval. That way the initial velocity is the same in each case, and you don't have to work out an initial velocity for leg2, the displacment would be OA + AB, and the time 2+3.
13. Again, I've learnt something new and I will try it out tomorrow. Perhaps tonight my head won't be trying to do the question without paper.
14. I have now completed the question taking on ghostwalker's advice to consider AB and AC rather than AB and BC. The whole question, including diagram, fits on one side of A4 rather than the many from previous workings.
15. i did it by assuming that suvat works and finding consistent values of u & a thus proving that the acceleration a is constant
16. Hi

As I'm only a beginner in this field of maths (mechanics), I'm keen to pick up tips. If it's not too arduous a task for you, could you post your full solution (with commentary for us beginners) to see if there's a different way to solving it than I have used. It's good to see other people's thought processes. I would never have thought of the way ghostwalker led me through it. (In fact, I would never have thought of a solution al all!!) Another view would be great.

Thanks in anticipation.
17. (Original post by the bear)
i did it by assuming that suvat works and finding consistent values of u & a thus proving that the acceleration a is constant
Well you can't prove that the acceleration is constant. Only show that the data is consistent with constant acceleration.

And your method is more or less what the OP did, is it not? Bar checking that the data is consistent with suvat at the end.

@OP This is the most unusual question I've seen for M1 - one I'd expect from MEI, which seems to be more challenging, at least based on the questions asked on here.
18. (Original post by ghostwalker)
Well you can't prove that the acceleration is constant. Only show that the data is consistent with constant acceleration.
yes you are right ghostwalker

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