A boat has a speed of 4 m s
−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
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.
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
To calculate the velocity of the boat relative to still water (u) and the velocity of the current (w), you can use the given speed and heading of the boat, as well as the speed and bearing of the current.
(a) To find u, you can use the boat's speed (4 m/s) and heading (170°). Convert the heading to radians by multiplying it by π/180. Then you can use trigonometric functions to calculate the x and y components of the velocity:
u_x = speed * cos(heading in radians)
u_y = speed * sin(heading in radians)
In this case:
u_x = 4 * cos(170° * π/180) ≈ 4 * (-0.985) ≈ -3.940 m/s (rounded to one decimal place)
u_y = 4 * sin(170° * π/180) ≈ 4 * (0.174) ≈ 0.696 m/s (rounded to one decimal place)
So the velocity of the boat relative to still water (u) is approximately -3.940 i + 0.696 j m/s.
To find w, you can use the current's speed (2.5 m/s) and bearing (105°). Convert the bearing to radians:
w_x = speed * cos(bearing in radians)
w_y = speed * sin(bearing in radians)
In this case:
w_x = 2.5 * cos(105° * π/180) ≈ 2.5 * (-0.259) ≈ -0.648 m/s (rounded to one decimal place)
w_y = 2.5 * sin(105° * π/180) ≈ 2.5 * (0.966) ≈ 2.415 m/s (rounded to one decimal place)
So the velocity of the current (w) is approximately 0.648 i - 2.415 j m/s.
(b) To find the resultant velocity of the boat (v), you can simply add the u and w vectors:
v = u + w
= (-3.940 i + 0.696 j) + (0.648 i - 2.415 j)
= -3.292 i - 1.719 j m/s (rounded to one decimal place)
So the resultant velocity of the boat (v) is approximately -3.292 i - 1.719 j m/s.
(c) To find the magnitude and direction of v, you can use the components calculated in (b).
The magnitude of v is calculated as:
|v| = √(v_x^2 + v_y^2)
= √((-3.292)^2 + (-1.719)^2)
= √(10.842 + 2.953)
≈ √13.795
≈ 3.714 m/s (rounded to one decimal place)
The direction of v can be determined using trigonometry:
θ = arctan(v_y / v_x)
= arctan((-1.719) / (-3.292))
≈ arctan(0.522)
≈ 27.6° (rounded to the nearest degree)
Since the boat's heading is measured clockwise from the North, we can find the direction of v by subtracting the angle from 360°:
Direction of v = 360° - 27.6°
≈ 332.4° (rounded to the nearest degree)
Therefore, the magnitude of the velocity v of the boat is approximately 3.714 m/s and the direction is approximately 332° (rounded to the nearest degree).