Applied Maths help!

roughage

Hi all,

I'm stuck on the relative velocity questions for 1998 and 1999 - part b of 99 especially I haven't a clue of. Any help much appreciated - thanks.

MathsManiac

Question 2 on 99 has no part (b). Is it part (ii) that's the problem, or are you talking about question 1?

roughage

No, it's definitely Q2 (b) part (i) that's causing the problem (I reckon I might even be able to part (ii) if I had part (i))


The question reads "Two ships A and B move with constant speeds 48 km/h and 60 km/h respectively. At a certain instant, B is 30 km west of A and is travelling due south. Find

(i) the direction A should travel to get as close as possible to B

(ii) the shortest distance between the ships"

MathsManiac

Sorry, I clicked on 1997 by mistake. Give me a minute.

roughage

Cool thanks - think I'm actually happy enough with the two part a's actually now - and think I might have the 98 part (b) using a diagram. But the shortest distance thing in '99 I haven't a notion of.

Katerrrs

1999 question 2b: see diagrams in attatchment

Diagram 1: Vab = Va -Vb

Vab = (-48cosAi - 48sinAj) - (-60j)

= (-48cosA)i + (60-48sinA)j

tanB = j over i = 60-48sinA over 48cosA
tanB = 5-4sinA over 4cosA

then differentiate tan B with respect to A (using quotient rule) and make it equal to 0

You get 16cos^2 A - 20sinA + 16sin^2 A

which simplifies to sin A = 4/5 , (and tanB = 3/4)

Then part 2: (from diagram 2)

|BX| = 30sinB
= 30(0.6)
= 18km QED

I'll put up 1998 if i get a chance!

roughage

Thanks a million Katerrrs. Had the second diagram completely wrong in my head - sorted now I think. Still finding 98 b a bit tricky so help would be great if you get a chance. Thanks a mill for that much though!

MathsManiac

Pipped at the post! (And I agree with Katerrrs' answer.)

Katerrrs

Okay, i've done my part. Now could someone please offer some help with question 10Bii on the 1995 paper? Can't come up with any substitution or other method of integrating that works on the 'v' side...anyone any ideas?

Cokehead Mother

Katerrrs wrote: »

Okay, i've done my part. Now could someone please offer some help with question 10Bii on the 1995 paper? Can't come up with any substitution or other method of integrating that works on the 'v' side...anyone any ideas?

My papers are in an awful mess so I don't know which questions are what year and stuff... is this the 150 - 3v / 2v one? If not, could you post what you're trying to integrate?

MathsManiac

(And the ones on examinations.ie don't go back that far.)

alan4cult

1998 Q2(b) attached.
My solution might be hard to follow!

Cokehead Mother

Wait, that was 1994. I have the 1995 one here now. I think.

You've got ma = mg - mkv

so a = g - kv

dv/dt = g - kv

int dv/(g-kv) = int dt

when v = 0, t = 0

When v = v, t = t.

To integrate the left, you say u = g - kv. Then du = kdv and then -du/k = dv.

So it integrates to -1/k ln(g-kv), and when you slosh in your limits you get

-1/k ln(g-kv) + 1/k lng = t

Then you clean that up to

ln(g - kv) = lng -kt

so g - kv = e^lng-kt

so v = g/k - 1/k e^lng-kt

terminal velocity is v as t tends towards infinity.

-1k e^(lng - kt) = e^lng / e^kt

As t tends towards infinity the denominator gets bigger so -1k e^(lng - kt) tends towards 0.

so the terminal velocity is just g/k.

I really hope that's the right question.

alan4cult

1995 Q10 attached.
For last part don't get complicated.
You know that a = g - kv

Speed will reach limit when it stops accelerating i.e. a = 0
a = g -kv
0 = g -kv
g = kv

v = g/k

Cokehead Mother

Are you allowed to just do that in the exam? The Oliver Murphy book does it the long way for some reason.

alan4cult

Cokehead Mother wrote: »

Are you allowed to just do that in the exam? The Oliver Murphy book does it the long way for some reason.

My App maths teacher said its fine but I'd would honestly do it your way to be safe. I'd only do it the short way if the examiner was ripping the answer book from me!