* Applied Maths * predictions / discussion / aftermath (1 thread only please)

ride-the-spiral

mr_brightside wrote: »

Does anyone know is there such thing as attempt marks, or is it the five or nothing? Is it in any way like the maths where one stupid slip only costs you a mark if all the actual maths bits are right?

Though Q6 was unreal, seemed a bit too good to be honest.. any one get omega = 7 ? And then 10 a i think i made a balls of, should it have worked out really evenly or was it sort of awkward?

Yeup, the applied maths marking scheme just doesn't emphasise it as much as the normal maths MS.

And to Brian, sorry but I can't actually remember and didn't write my answers down, all i remember was they weren't very "nice".

KoolAidRelic

A slip is -1, and a blunder is -3 - just like normal maths.

I got omega as 7 as well.

mr_brightside

Also in the linear motion, the shortest time i got was (11t/6), which doesn't make sense cause that'd be shorter than the original journey which was at a greater speed... anyone get it out?

And just one more (i swear), alpha in q3(a)... 75. something anyone?

Athos902

I thought it was quite nice overall.

Did anybody else get e=-0.75 for Q5 (b) (ii)?

Geog ariphic

mr_brightside wrote: »

.. any one get omega = 7 ? And then 10 a i think i made a balls of, should it have worked out really evenly or was it sort of awkward?

Got c=7, i think it was all logs.

anna_k

Athos902 wrote: »

I thought it was quite nice overall.

Did anybody else get e=-0.75 for Q5 (b) (ii)?

i'm probably completely wrong here but surely e must be positive?
(btw, i can't remember what i got for e exactly...some nice positive fraction)

john1741

Athos902 wrote: »

I thought it was quite nice overall.

Did anybody else get e=-0.75 for Q5 (b) (ii)?

I got +0.75, take the absolute value because 0<e<1

BL1993

I can confirm that these are the correct answers for question 10:

Q10) a) y=66.231
b) i) 44.629m
ii) 0.336 ms-1

japester

I could be wrong but I've just done Q10 there and I get the same results as you for (a) and (b)(i) but for b(ii) I am getting 16.249m/s which would seem about right given that u = 40m/s and v = 0m/s (in the case of a uniform deceleration, we would expect the average speed to be 20m/s). Your value (I think!!) is way too low.

Geog ariphic

@BL1993, that last one looks VERY familiar - if i was a gambling man i'd say that was what i had down for average velocity. Did u just divide max distance by total time aswell? xD

BL1993

@japster: did you intergrate again using dv/dt and not vdv/dx? Because you will be getting a tan inverse and not a log for the speed side. Well it may appear to be low but the maths doesn't lie. If the time it takes to reach 0 is big then the average speed is going to be low.

@Geo: Yes, it's the only way it can be done

electrictrad

Athos902 wrote: »

Did anybody else get e=-0.75 for Q5 (b) (ii)?

anna_k wrote: »

i'm probably completely wrong here but surely e must be positive?
(btw, i can't remember what i got for e exactly...some nice positive fraction)

Got a nice positive fraction too. . .7/8. . .I'm probably wrong though. . .

BL1993 wrote: »

I can confirm that these are the correct answers for question 10:

Q10) a) y=66.231
b) i) 44.629m
ii) 0.336 ms-1

I got a) y = 2e^3.5
b) i) 608m approx
ii) around 4.3 m/s

. . .so I hope to God you're not the SEC. . .

mr_brightside wrote: »

Also in the linear motion, the shortest time i got was (11t/6), which doesn't make sense cause that'd be shorter than the original journey which was at a greater speed... anyone get it out?

And just one more (i swear), alpha in q3(a)... 75. something anyone?

I got shortest time as 31t/12, and x=50 for Q3a) i). . .how did people do ii)?. . .I said sy/sx must be tan 45 (as i=j for my part i) and worked ut out that way. . .probably wrong though

japester

@ BL1993, I've just checked my calculation there for Q10 (b) (ii) and I do see that I made a slip! I did the integration alright to find the time for the total deceleration and in my corrrect (??) version I am now getting 2.318s for the time and therefore 19.253m/s for the average speed. I completely appreciate what you are saying about the fact that it could take a long time for the deceleration and therefore average speed could be very low.

Geog ariphic

electrictrad wrote: »

Got a nice positive fraction too. . .7/8. . .I'm probably wrong though. . .

ok i lied earlier, i DO remember getting 7/8 now, for e. Allow me to go compltely wild and say whoot. *Whoot*

Bte ppl are giving their answers as 200lnsomething and e to powers, and they usually have it like that in marking schemes, but you can just work it all out to decimals and get ur marks right?
I know correctors are usually fair but i don't trust anything to do with correcting applied maths. Some of it is so off the wall.

japester

@BL1993 I've just spotted where our answers differ in Q10(b)(ii), when you are getting arctan(40/80), you are getting the angle as 26.56 degrees. However, I am converting this angle to radians to get 0.4636 radians. This accounts for the different results we are getting. Any past questions that I have come across involving arctan appear to expect the angle to be expressed in radians - its a dirty little twist in the question really though

Geog ariphic

They cannot hold you for not giving your angle in particular units when they do not specify and you just go with whats normal for angles, SURELY?

BL1993

@electricrad: For part (a) your answer is the same as mine if you fully evaluate it. However, I don't believe that your second part is right. I'm pretty sure that my first part is right for (b) but now that I read japster's post, I believe that he is in fact right in saying that the angle must be converted to radians. Thanks japster @Geo: Well when dealing with tan or sine inverse, I always had to use radians to solve problems so i don't see why it wouldn't be the case here. Well spotted japster.

Geog ariphic

You always 'had to' use radians to solve? why?
Alright, whenever there is an arcsine or arctan formula to solve it always equals constant times pi. But just beacuse you are finding the angle, if u have say tan(theta) = 8, does not mean you have to give the answer in radians.
Does anyone kno, have they actually penalised this before?

BL1993

Because it is the S.I unit of angular measure

Gyaradose

ANy idea on what the hell question 3 (a) was? if you rounded the values of sine and/or tan or even the initial speed the quadratic cannot be done, i.e. b(squared) - 4ac is negative, but using the true value for the initial speed b(squared) - 4ac =0

japester

Not many people attempt Q9 but for anyone that did, part (a) was very reasonable (U-tube) and part(b) took a bit of thinking about but still fairly doable. The answers I got for this Q are (a) 1.44cm3 and (b) 3/7. Just be interested to know if they tally with anyone who tried it out.

Gyaradose

japester wrote: »

Not many people attempt Q9 but for anyone that did, part (a) was very reasonable (U-tube) and part(b) took a bit of thinking about but still fairly doable. The answers I got for this Q are (a) 1.44cm3 and (b) 3/7. Just be interested to know if they tally with anyone who tried it out.

I got the same for part a but my part b was dreadfull so I cant confirm that

anna_k

japester wrote: »

Not many people attempt Q9 but for anyone that did, part (a) was very reasonable (U-tube) and part(b) took a bit of thinking about but still fairly doable. The answers I got for this Q are (a) 1.44cm3 and (b) 3/7. Just be interested to know if they tally with anyone who tried it out.

I got 3/7 for 9b) too! yay!!...part a) on the other hand...it was the last question i did ( i did b before a for some reason) so i was running out of time...god only knows what stupid answer i put down..but i'm in no doubt that it was wrong...and u-tubes are so easy :mad:

japester

I've just done Q1 there and the answers I'm coming out with seem to match with those of redredwine earlier. They are (a) 0.875 s (b) (i) speed-time graph (ii) t1+t2 = t and (iii) 5t/3. I'm not 100% sure of the answer I'm getting to (iii) however, because I am getting an equation for s that uses the acceleration and deceleration from part (ii) (as instructed) to get the expressions for these parts of the journey in terms of t but then I am saying that the distance traveled for the "flat" part of the journey is (2/3)vt which is not the case surely (the t in this part is not the same "t" from part (ii) i.e. it will have to be greater in part (iii) to compensate for the fact that the distances covered in the acceleration and deceleration phases are now lower and the total distance covered must be a constant. Seriously confused with this part Maybe someone can shed some light on it - I know I came across one like it somewhere before but I probably made a mess of it too then!!

Geog ariphic

japester wrote: »

(iii) however, because I am getting an equation for s that uses the acceleration and deceleration from part (ii) (as instructed) to get the expressions for these parts of the journey in terms of t but then I am saying that the distance traveled for the "flat" part of the journey is (2/3)vt which is not the case surely (the t in this part is not the same "t" from part (ii) i.e. it will have to be greater in part

Yes ur right you cannot use the same t values. i called them t1, t2, t in the first situation, ta, tb, tc in the second (ta+tb+tc=T).
However i never used the acceleration, so ended up with T(+/-)tb=t (i forget whether it was plus or minus)

so maybe u can make something with that, im not sure.

BL1993

I am a million percent sure(did question out 3 times) that the answer to q.1 part 3 is 2t. let T1, T and T2 equal the times that the body accelerates, moves at constant velocity and decelerates respectively. The acceleration is equal to the slope of the graph if you drew it out. I.e, a=v/t1 which is equal to 2v/3T1`and you rearranged it until you get T1=2t1/3. Do the same for the deceleration. Then you find the distance it travels in case 1 and 2 and let them equal, mess around with the equation until you get T=2t. I cba arsed typing it out but I hope you guys see the point. Remember to bear in mind that t1+t2 = t

japester

@ BL1993 - I see what you're doing alright and I was getting the acceleration and deceleration times in a similar manner, with reference to the earlier situation. I'm getting the acceleration time for the new situation as (2/3)t1 and similarly for the deceleration, it will be (2/3)t2. I'm happy with this much of it and the distance can be determined for these parts. As I said earlier, getting the time that the body moves with constant velocity takes a fair bit of effort and I'm basically saying that the distance covered must be the same in both cases. I'm using the speed-time graph in the original situation to get the area and then doing likewise for the second situation. I am letting the time spent traveling with constant speed in the second situation be T and this must be > t from the first situation in order that the body covers the distance (as the speed limit is lower). In my equations here I am getting terms in t1, t2 and T, which I can then solve for T, provided that I leave (t1+t2) be equal to t. And this is where I have my serious doubts I believe that the expression t = t1 + t2 is true only provided that the average speed is 3v/4, as it was in the original case. In the second case, the average speed cannot be 3v/4 (it must be lower) and therefore how can I let t = (t1+t2) in my equations? I'm probably making some basic error here but I can't figure this one! For what its worth, if I do actually assume that t = (t1 + t2) is valid, then I am getting a value of T = (23/12)t for the constant speed in the second situation and then I end up with 3t/2 for my answer. Its a tricky little bugger

japester

@ geographic, thanks for the advice on that question - as I mentioned in my last post, its a tricky little one and I think that its possibly simpler than I am giving it credit for I can't imagine that it could be worth more than 10 marks, given that (i) and (ii) take a bit of time themselves and therefore I suspect that the solution is more straightforward than I think!

Geog ariphic

You're right there i'd say, however, and i may be reading this wrong, but u seem to have confused time notation from the first scenario and the second?

And yes i did get t1+t2=t in the first part, but if u are calling any parts of the time in ur second scenario t,t1 or t2 (which u cant since they were all used and all three times are now different), then no it would not be the same.

BL1993

Times in case one: t1, t, t2
Times in case 2: T1, T, T2

I probably shouldn't have sued capitals and should have called them x, y and z :P sorry about that.