someone kelp me please!!!!

- YumYum247

someone kelp me please!!!!

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- schrodinger

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- YumYum247

Please check my second question.....

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- YumYum247

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- YumYum247

please check my method and calculations with a calculator........

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## More answers

- anonymous

Do you not have a calculator?

- YumYum247

is my method correct?????O_O

- YumYum247

@IrishBoy123 can you please check my work??????:)

- anonymous

Part A)
Her velocity relative to the short is equal:
\[v=\sqrt{(1.2\frac{km}{h})^{2}+(2\frac{km}{h})^{2}} \approx 2.3\frac{km}{h}\]
Which by my calculation should give you an angle of:
\[\tan^{-1} (\frac{ 1.2 }{ 2 }) \approx 31^{o}\]

- anonymous

The method you choose for part b seems correct... using the known velocity across the river coupled with the width of the river to find the travel time.... then using that travel time to calculate the vertical distance downstream the boat has gone.

- YumYum247

t your part one doesn't look correct......hold up let me upload the first question....:)

- anonymous

However for the time i found 46.08s... but this problem is using 2 significant figures so 46s is probably sufficient

- YumYum247

it's because i think the numbers that your inputting aren't so accurate....

- anonymous

Crap your right I was using the wrong horizontal velocity (2 km/s) :( lemme recalc

- YumYum247

let me upoad the first question, may take some time....

- anonymous

Ok I got 2.8 (rounded from 2.77) km/s and 26 degrees (rounded from 25.6)

- YumYum247

yes 26 is what i got.

- anonymous

so for part b like i was saying the idea is correct, but i got different numbers (see above) that I did use the correct velocity to calculate (2.5 not 2)

- YumYum247

try plug in those new numbers....

- anonymous

so in 46 seconds the boat travelled downstream a distance d=vt=(1.2*1000/3600) m/s * 46s = 15.33s or roughly 15 seconds (again 2 sig figs.... I dont know if your teacher is a stickler for that)

- anonymous

darn i meant meters... 15.33 or just 15 m

- anonymous

I did YumYum i plugged in the right numbers and am getting slightly different answers in my calulations

- YumYum247

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- anonymous

Ok for part c) its a little bit more involved but still not too bad hold on....

- anonymous

Yes YumYum i posted those corrected numbers above

- YumYum247

let me upload my part c.....

- anonymous

so basically speaking part C) is looking for you to correct for the downward drift by aiming the boat upwards

- YumYum247

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- YumYum247

please check my east and north signs, are they correct? at the bottom of the page!!!

- YumYum247

i think it should have been (N29E), what do you think.????

- YumYum247

don't worry about medals, your about to get plenty from me ^_^

- anonymous

So you correctly note the hypoteneuse is of magnitude 2.5 km/s because that is max speed of her boat... which gives you an angle of sin^-1(1.2/2.5) ~28.68~29degrees

- YumYum247

i wrote 29degrees!!!

- anonymous

then you found the horizontal velocity to be 2.5^2=v^2+1.2^2 or v=sqrt(2.5^2-1.2^2)~2.19~2.2km/h

- YumYum247

i don't know why i wrote 2.2"KM/h"

- YumYum247

it should have been 2.2Km up in the north.....

- anonymous

as for the formatting of the direction.... that I am a little unsure about since it has been a while... but your angle is correct so i have to leave that to you unfortunately :(

- YumYum247

does it all look right?!?!?!?!?

- anonymous

so if you take north as your main direction you would be heading at an angle 90-29=61degrees east from the north or 29degrees north from the east... something like that

- anonymous

So yea i agree it all looks good... so far so good

- YumYum247

aight thanks!!! Please come online more often, i might need your help.....24/7 :"D
Thanks!!!!

- anonymous

now part d is confusingly worded in my opinion but if I understand it correctly than it is easy... just reproduce what you did in part b using your new horizontal velocity to calculate the time to travel across the river here

- anonymous

since she will arrive directly at the farmers market that will be the TOTAL travel time to the farmers market using this route

- YumYum247

it ok, i'll ask irishboy123....chill out

- anonymous

ok i looked back up above and in the first possible path she ended up approx 15m to the south of the market

- anonymous

so use her walking velocity to calculate the time it will take her to walk that distance north to the farmers market..... add it to the time it took her to cross the river... and that will give you the TOTAL time of the trip using this route

- anonymous

The conclusion should be clear at that point which is the best route to talk (quickest)

- anonymous

take*

- YumYum247

Thanks!!

- anonymous

Your welcome :D

- YumYum247

don't mind but i need to confirm with irishboy ^_^

- YumYum247

cuz 2 is better than 1 :D

- YumYum247

i do it all the time....:)

- anonymous

yea no probs i have seen irish boy he is good.... i couldve sworn i saw you two working earlier but when i saw this post again there was no content which is why i jumped in

- YumYum247

take it easy man!!! ^_^
come more often, we need people like you here on OS!!! :)

- anonymous

Yea I do when I have some spare time, but i am in university currently and have my own work and research to be doing.

- YumYum247

Thanks for all your help!!!! you sound very very smart tho!!! :)

- YumYum247

understood!!!

- YumYum247

Alright Thanks for everything, i hope i see you more often ^_^ Cheers!!!

- anonymous

Thank you :D and you are very welcome I will when I can

- YumYum247

Yes!!!!! :D

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