A boat has a speed of 4 m s

Question

−1
(relative to the surrounding water) and a
heading of 170◦
. There is a current flowing from a bearing of 105◦ at a
speed of 2.5 m s
−1
. Take i to point east and j to point north.
(a) Express the velocity u of the boat relative to still water and the
velocity w of the current in component form, giving numerical values
in m s−1
to one decimal place.
(b) Express the resultant velocity v of the boat in component form, giving
numerical values correct to one decimal place.
(c) Hence find the magnitude and direction of the velocity v of the boat,
giving the magnitude to one decimal place in m s
−1 and the direction
as a bearing to the nearest degree

So far, I've got -

Part (a):
boat velocity vector u = 4 Cos(170o) i + 4 Sin(170o) j (relative to still water)
u = 4 (-0.985) i + 4 (+0.174) j
u = - 3.940 i + 0.696 j m/sec

current velocity vector w = - 2.5 Cos(105o) i - 2.5 Sin(105o) j (relative to still water)
w = - 2.5 (- 0.259) i - 2.5 (+0.966) j
w = + 0.648 i - 2.415 j m/sec

Part (b):
Resultant boat velocity vector V = u + w
= [ - 3.940 i + 0.696 j ] + [+ 0.648 i - 2.415 j ]
= - 3.292 i - 1.719 j m/sec

Part (c):
Magnitude of vector V = [(-3.292)2 + (-1.719)2] (1/2) = √(13.792) = 3.714 m/sec
Direction of resultant V: Sin Θ = -1.719 / 3.714 = - 0.463
Θ = 207.6o

Answer 1

Seems like a lot to do about nothing....
I made a sketch where OA is the vector of the boat in still water (ending up in quad II)
and OB the current vector, ending up in quad IV
I then completed the parallelogram ACBO.
With some simple angle calculations, I found
angle CAO = 65°
our resultant is OC, and by the cosine law

OC^2 = 4^2 + 2.5^2 - 2(4)(2.5)cos65
= 13.7976..
OC = 3.7145 , which is what you had

by the sine law:
sin AOB/2.5 = sin65/3.7145
sin AOB = .6099769..
Ø = 37.5897

resultant angle = 180 + (37.5897 - 10) = 207.59° , again your answer

compare your calculations with mine.

Answer 2

Hi there

From a wild guess are you undertaking TMA02 from MST 124 with the Open University?

Thought so.......me too!

Don't take this as gospel.....please....

I have an issue with this question, but as far as your answer goes I have a different answer to part a) - I think the boat is heading in the opposite direction to j - therefore j must be a negative - I think you may have to revisit the angle of 170 - in the fourth quadrant (ASTC rule)

As I'm doing the same course I wouldn't want to go further as it may breach strict OU rules regarding own work/plagarism etc