anyone worked out Q7 (a) (iii)? (1 Viewer)

[davidspy]

when does the ship leave?

 

[Sweet_Lemon]

didn't attempt.... tat question.. saw it somewhere...still can't figure out!

 

[Jase]

everyone's getting 4:28am

 

[Giant Lobster]

Omg I Love Yo Guys 4:28 Am Omgs
Wfuckan!!~!mkq#hr$iuh

 

[acmilan]

I found it to be 5:30 am but i could be wrong

 

[Jase]

yah i actaully found it to be 5:35.. but the majority seems to be getting 4:28 .. i dont know how though

 

[~ ReNcH ~]

Haha....didn't even attempt it
I reckon I would have got about 6/12 for Q7

 

[bertevans]

the time value i got for 7 was 4:28, but isn't that the time after 1:00am that the ship must leave, hence making the leaving time 5:28

 

[mgorman]

i got 6.40.
i just let t = t-1 for the harbour entrance, since whatever is happening at the wharf is just happening at the harbour entrance an hour earlier, and then set y =6 since the tide has to be greater than 6. The tide will already be greater than 6 since it starts at a high tide at 2am, and hence the value you get for y =6 will be the last possible time that the ship can get through. this gives you 4.801 (using radian mode in calculator), which equates to 4 hours and 48 minutes, 6:48am. But ship has to leave at least 20 mins before 7am, hence it can leave at 6.40 as the LATEST possible time. at 6.40, the tide at the wharf will still be less than 8.5m so i can't see how this doesnt work.
Open to any suggestions.

 

[cameron0110]

I got 4:28 am, but I can't really remember how I did it, but it made sense at the time.

 

[Shuter]

I'll work it out after my eco test tommorow. I didn'thave time so put down a blank guess of 6:35am.

 

[Rorix]

i got 6.40.
i just let t = t-1 for the harbour entrance, since whatever is happening at the wharf is just happening at the harbour entrance an hour earlier, and then set y =6 since the tide has to be greater than 6. The tide will already be greater than 6 since it starts at a high tide at 2am, and hence the value you get for y =6 will be the last possible time that the ship can get through. this gives you 4.801 (using radian mode in calculator), which equates to 4 hours and 48 minutes, 6:48am. But ship has to leave at least 20 mins before 7am, hence it can leave at 6.40 as the LATEST possible time. at 6.40, the tide at the wharf will still be less than 8.5m so i can't see how this doesnt work.
Open to any suggestions.

if the harbour enterance is t=t-1, then when something is happening at the wharf, it happns an hour later at the entrance

it should be t=t+1
i made the same mistake

 

[superbird]

the ship will leave soon

 

[dafidav]

if you wrote 4.28am then isnt that for the earliest time for the ship to leave??

it asked for the lastest time...

 

[:: ck ::]

fck i did that stupid mistake forgot abt the hr ARGHHHHHH

 

[skankit]

did anyone else get the time as like 5:19am??????

damnit, i totally thought id caned that question...

 

[Gesus]

if you wrote 4.28am then isnt that for the earliest time for the ship to leave??

it asked for the lastest time...

Heres the working for it:
You dont really need i or ii for it, just the cos bit.

Let 4(pi)t/25 = X
High tide - Low tide = 6m (given)
Hence Amplitude = 3
water must be at least 2m above low ( 1 below equilibrium)
hence
-1=3cosX
X=arccos(-1/3)
4pi x t/25 = arccos(-1/3)
t = 25arccos(-1/3) / 4pi
t = 3 Hr 48 min
but, tides occur 1 hour earlier than the wharf, so t-1 = 2Hr 48 min after high tide @ 2a.m. (wharf)
so due to the nature of the Cosine curve, the first answer is the earliest (and the 7am condition), so ship needs to be at the habour entrance by 2am + 2.48 = 4.48.
20 minute trip from wharf to habour, so latest time = 4.48 - .20 = 4.28am


It can only leave between 4.05am and 4.28am, hence 4.28 is latest

 

[beverly]

i got 5.35 as well i think..

 

[Kirsti]

Gesus makes me feel good bout my answer I thought sure I'd fucked it but I'm feelin better now