Leaving Cert Applied Maths Thread

Lucy Peden · 02-02-2018 11:33PM

Account Number wrote: »

Well lads, this question appeared on our mocks, and it seemed to cause a bit of confusion.
Assuming I did it right:
s=ut + 1/2(at^2) ---- 240 = 16t --- t = 15
s=ut + 1/2(at^2) ---- 270 = (19)(15) + 1/2(225a)
Essentially a is a minus answer, and surely this can't be right.Someone care to show me the error of my ways?

I attempted it and got a=-2/15

bpb101 · 02-02-2018 11:47PM

I haven't done applied maths for years and I attempted it. Got 64km/h/h ðŸ˜ðŸ˜‚ðŸ˜‚ðŸ˜‚ðŸ˜‚

bpb101 · 02-02-2018 11:56PM

Right. Done it again and I get a reasonable answer. Just over 1ms/s

bpb101 · 03-02-2018 12:05AM

Woops I made a simple mistake.
Maybe the guy doing it done the same. I got a minus answer of -0.138. But this could be right.
The horse could have been catching up but easyed off. But it is misleading.

Lucy Peden · 03-02-2018 12:08AM

bpb101 wrote: »

Right. Done it again and I get a reasonable answer. Just over 1ms/s

I found the time at which mojo passes the finish line and worked out the answer from that, he passes 10 metres before starlight and so it work out to be 15 seconds, then I used s=ut + 0.5at^2 and solved for a

ignorance is strength · 03-02-2018 02:16PM

As Lucy said, answer is -0.13 m/s^2, which is got using the method you give. Not sure why you think a negative value is illogical. If you visualise the race on a velocity-time graph, despite a decreasing velocity, "Mojo" remains faster than "Starlight" throughout.

Account Number · 03-02-2018 02:46PM

Thanks, lads -2/15 was the value I got as well, I just thought it wasn't right because the question said Mojo sped up, implying a positive acceleration. I guess I just gave it a bit toooo much thought..:P Thanks anyway

ignorance is strength · 03-02-2018 02:59PM

Account Number wrote: »

Thanks, lads -2/15 was the value I got as well, I just thought it wasn't right because the question said Mojo sped up, implying a positive acceleration. I guess I just gave it a bit toooo much thought..:P Thanks anyway

I hadn't noticed that. I think you're right and it's a mistake in the question.

bpb101 · 03-02-2018 03:12PM

Account Number wrote: »

Thanks, lads -2/15 was the value I got as well, I just thought it wasn't right because the question said Mojo sped up, implying a positive acceleration. I guess I just gave it a bit toooo much thought..:P Thanks anyway

i would say they mixed up mojo and starlight's speeds

ignorance is strength · 03-02-2018 03:38PM

Easily done. Mojo's got some speed on him.

But I imagine it might have happened during the editing stage of the question -- either the figures were retrospectively changed so as to make the acceleration negative, and the verb wasn't changed, or the verb was added by someone not familiar with the answer. It might have read something like, "Mojo catches up with an acceleration...", and that, in the edit, that was seen to be imprecise because Mojo continues the negative acceleration after the overtake. (Obviously just speculation.)

Lucy Peden · 03-02-2018 05:59PM

Did anybody manage to show in the projectile question that the initial speed was 28.6m/s? I ve been trying to get it out but must be missing something

ignorance is strength · 04-02-2018 01:32AM

Lucy Peden wrote: »

Did anybody manage to show in the projectile question that the initial speed was 28.6m/s? I ve been trying to get it out but must be missing something

Will you post the question?

Lucy Peden · 04-02-2018 11:47AM

ignorance is strength wrote: »

Will you post the question?

The first part of that

Lucy Peden · 04-02-2018 11:48AM

Don't think it sent

ignorance is strength · 04-02-2018 02:03PM

I think it’s fairly standard. You’re probably just making a trivial error somewhere.

Sx = ucosAt = 52.8, from which you get t = 57.2/u. And then inputting that value for t into Sy = usinAt - gt^2/2 = 2.4 gives you the answer for u.

The second part is somewhat unclear because in reality the answer ought to be the range of values for x between the two times the ball’s height equals the goalkeeper’s reach. It also assumes that any contact between the keeper and the ball will prevent a goal.

randomuniquser · 12-02-2018 04:57PM

How did everyone find the Applied Math mock ?

I personally didnt think it was too bad but q6 (a) wording was quite difficult to understand.
So I chose to do q5 instead.

q9,q10,q2,q1 were pretty easy but 3 part b and 5 part (b) (i) i found a bit tricky.

Overall the paper was pretty fair I think but I don't think it accurately represents what is going to come up in the LC.

Does anyone else feel the same?

Account Number · 07-05-2018 08:53PM

Anyone care to point out where I'm forgetting to carry the one?:p
2017 Paper Q3 (a)(i) I have up to:
(9gh/2)^1/2sin(a)(3h/(9gh/2)^1/2 cos(a))- g/2 (3h/(9gh/2)^1/2 cos(a))^2 (Sorry that's probably one of the ugliest things ever to read), and I can't seem to get the right answer out of it. I keep ending up with tan^2(a)-12tan(a) +3, which doesn't get me the right answer.The marking scheme subs in u for the speed, but I can't make out where it's coming from

skippy1977 · 08-05-2018 05:15PM

It seems to be right...I used the same method, it might just be the way you are tidying up? Here is what I have if it is any help.

lostatsea · 08-05-2018 09:23PM

skippy1977 wrote: »

It seems to be right...I used the same method, it might just be the way you are tidying up? Here is what I have if it is any help.

Congratulations on your beautifully presented solutions.

Account Number · 05-06-2018 08:03PM

Hello, me again. Quickly trying to cover a final topic, in the run up, and am in the middle of Q8. Anyway came across a q in the Oliver Murphy book, which I can't seem to make sense of.Ex 14B Q7 7(ii) I understand it's a form of potential energy, but I can't work out where the h values are coming from.Sorry this is probably a stupid question, but I can't work it out for the life of me:o and I couldn't find a video from our friend on YouTube on it:o

bpb101 · 05-06-2018 09:25PM

Account Number · 05-06-2018 10:09PM

bpb101 wrote: »

Thanks, much appreciated,

but I meant 14B with the uniform rod PQ

. I have the solution for it, I'm just trying to work out where they're getting the values for h1 and h2 from, in the formula.:o

bpb101 · 07-06-2018 09:23PM

Been so long, i forgot there was parts to the chapters

I have all the solutions worked out, so give us a shout if you want a particular answer or pm me and i can send them to you

Celtron · 20-06-2018 09:39PM

What questions are ye all doing? I've covered 1,2,3,4,5,6 and 10, been told 9 is very doable, should I bother starting it now or would I just be wasting time?

lundrum · 20-06-2018 09:43PM

We're doing 1,2,3,4,5,9 and 10. I'm not sure if there's enough time to start on 9 now but fwiw I think it's among the easiest of the questions we've done most years.

Thr000w · 21-06-2018 12:41PM

We've done 1 through 6 and 10.

lundrum · 22-06-2018 05:09PM

What did everyone think of that? My H1 is down the drain I'd say

Curiouscircus · 22-06-2018 05:21PM

lundrum wrote: »

What did everyone think of that? My H1 is down the drain I'd say

Same, no hope of my H1 after that. V difficult. Ended up running out of time as well.

lundrum · 22-06-2018 05:24PM

Curiouscircus wrote: »

Same, no hope of my H1 after that. V difficult. Ended up running out of time as well.

Which questions did you do? I did 1,3,4,5,9 and 10 and a really pathetic attempt at 2 that was basically just a waste of time

Curiouscircus · 22-06-2018 05:26PM

lundrum wrote: »

Which questions did you do? I did 1,3,4,5,9 and 10 and a really pathetic attempt at 2 that was basically just a waste of time

I did 1,2,3,4,5,10

Oh yeah like I read part b of question 2 completely wrong. Misinterpreted it so badly I don’t even think I got any marks for it
Questions 10 and 3 were nice enough though

*Leaving Cert Applied Maths Thread*

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Leaving Cert Applied Maths Thread