A river is flowing due east at 8 mi/hr. A man heads his motorboat in the direction N30E in the river. the speed of the motorboat relative to the water is 12mi/hr a)express the true velocity of the motorboat as a vector b) find the true speed and direction of the motor boat.

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A river is flowing due east at 8 mi/hr. A man heads his motorboat in the direction N30E in the river. the speed of the motorboat relative to the water is 12mi/hr a)express the true velocity of the motorboat as a vector b) find the true speed and direction of the motor boat.

Mathematics
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@zepdrix free? (:
@jim_thompson5910 whenever you're free, please! :)
|dw:1434672450418:dw|

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Other answers:

Draw in a parallelogram |dw:1434672391987:dw|
this vector, ie the resultant vector, is the boat's true path |dw:1434672449672:dw|
oo ok
Following so far
and the resultant is w = u + v right?
what is the vector component form for the vector pointing directly east (the river current)?
emm v = 0i + 8j ?
close
that's pointing north though
oops sorry v= 8i + 0j
correct
what is the boat's vector (ignore the current) ?
hint: |dw:1434672891389:dw|
Um..not sure
z = x*i + y*j x = r*cos(theta) y = r*sin(theta) theta = 60 degrees r = 12
x = 6 y = (12 sq3)/2
or y = 6*sqrt(3)
or y = 6 sq3
yep
haha yes
so that's the boat's vector <6, 6*sqrt(3)>
|dw:1434673336710:dw|
|dw:1434673363432:dw|
so is it = <14, 6 sq3>
yes, that's u+v
which is also the true velocity vector of the boat
yaaay! Now part b :)
true speed = |u+v| true direction = direction of vector u+v
Direction = 2 sq76 ?
or 4 sq19
4*sqrt(19) is the magnitude of u+v so it is the true speed
o right. Direction: (theta) = arctan( (6sq3)/14)
so.. 36.58?
but if we want to find it from x it would be that subtracted from 180 correct?
sorry, I meant 90
Which would be 53.41
N53E
yeah I'm getting theta = 36.58677555 so 90 - theta = 90 - 36.58677555 = 53.41322445
Yay, it matches :)
a) velocity of motorboat: <14, 6*sq3> b) Speed: 17.44 Direction: N53.4E
everything looks good
thank you
yw

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