Answers
I believe you need the velocity of the speedboad in order to solve this. (The timeframe to interception would be satisfactory.)
The approach will be to get the ship's location as a function of time. Initially,
y = 21km * cos15º = 20.3 km
x = 21km * sin15º = 5.44 km
y(t) = 20.3km + 22km/h * cos40º * t
x(t) = 5.44km + 22km/h * sin40º * t
assuming you had the speedboat's speed, you could create similar eqn's of motion:
Y(t) = ....
X(t) = ....
both also functions of some unknown angle T
You need to have a time t such that
y(t) = Y(t) and
x(t) = X(t)
.
The unidentified ship starts at a distance of 20 km, at 15º east of north,
with a velocity of 26 km/h, at 40º east of north.
These (r,?) polar coordinates can be transformed (x,y) coordinates by using the following formulas.
x = r sin?
y = r cos?
So the ship unidentified starts at position ( 5.18x + 19.3y ) km
with a velocity of ( 16.7 x + 19.9y) km/h
The speedboat starts at corrdinates (0,0), and has an initial velocity of (50.0 , f) in polar coordinates. In cartesian, this becomes ( 50sinf , 50cosf)..
At a given time t from the start, the x-coordinate of the ship will be 5.18 + 16.7t, while that of the speedboat will be 50sinf*t.
The y-coordinate of the ship will be 19.3 + 19.9t while that of the speedboat will be 50cosf*t.
We want the solution for which the equations
5.18 + 16.7t = 50sinf*t
19.3 + 19.9t = 50cosf*t
are both satisfied. This is a system of two independdant equations with two unknowns, which can be solved.
Taking the first equation, we have
t = 5.18 / (50sinf - 16.7)
Plugging it into the second one gets us
19.3 = (50cosf - 19.9)t
19.3 = (50cosf - 19.9) * 5.18 / (50sinf - 16.7)
3.732 = (50cosf - 19.9) / (50sinf - 16.7)
3.732 = (cosf - 0.398) / (sinf - 0.334)
3.732 * (sinf - 0.334) = (cosf - 0.398)
3.732 sinf - 1.247 = cosf - 0.398
3.732 sinf = cosf + 0.849
Let's take the square of each side
13.93 sinf^2 = cosf^2 + 1.70 cosf + 0.721
Knowing that [sinf^2 + cosf^2 = 1], we have
13.93 (1 - cosf^2) = cosf^2 + 1.70 cosf + 0.721
13.93 - 13.93 cosf^2 = cosf^2 + 1.70 cosf + 0.721
14.93 cosf^2 + 1.70 cosf - 13.21 = 0
This is a simple quadratic equation in cosf.
The solutions are
cosf = 0.885
cosf = -0.999
acos (0.885) = 27.7º
acos (-0.999) = 177º (clearly not what we're looking for)
So it seems the direction in which to head is about 27.7º east of north
As in the start, this direction can be expresed in (x,y) coordinates by using
x = r sin? ; y = r cos?
so it becomes; v = (23.2x + 44.3y) km/h.
The unidentified ship starts at a distance of 20 km, at 15º east of north,
with a velocity of 26 km/h, at 40º east of north.
These (r,?) polar coordinates can be transformed (x,y) coordinates by using the following formulas.
x = r sin?
y = r cos?
So the ship unidentified starts at position ( 5.18x + 19.3y ) km
with a velocity of ( 16.7 x + 19.9y) km/h
The speedboat starts at corrdinates (0,0), and has an initial velocity of (50.0 , f) in polar coordinates. In cartesian, this becomes ( 50sinf , 50cosf)..
At a given time t from the start, the x-coordinate of the ship will be 5.18 + 16.7t, while that of the speedboat will be 50sinf*t.
The y-coordinate of the ship will be 19.3 + 19.9t while that of the speedboat will be 50cosf*t.
We want the solution for which the equations
5.18 + 16.7t = 50sinf*t
19.3 + 19.9t = 50cosf*t
are both satisfied. This is a system of two independdant equations with two unknowns, which can be solved.
Taking the first equation, we have
t = 5.18 / (50sinf - 16.7)
Plugging it into the second one gets us
19.3 = (50cosf - 19.9)t
19.3 = (50cosf - 19.9) * 5.18 / (50sinf - 16.7)
3.732 = (50cosf - 19.9) / (50sinf - 16.7)
3.732 = (cosf - 0.398) / (sinf - 0.334)
3.732 * (sinf - 0.334) = (cosf - 0.398)
3.732 sinf - 1.247 = cosf - 0.398
3.732 sinf = cosf + 0.849
Let's take the square of each side
13.93 sinf^2 = cosf^2 + 1.70 cosf + 0.721
Knowing that [sinf^2 + cosf^2 = 1], we have
13.93 (1 - cosf^2) = cosf^2 + 1.70 cosf + 0.721
13.93 - 13.93 cosf^2 = cosf^2 + 1.70 cosf + 0.721
14.93 cosf^2 + 1.70 cosf - 13.21 = 0
This is a simple quadratic equation in cosf.
The solutions are
cosf = 0.885
cosf = -0.999
acos (0.885) = 27.7º
acos (-0.999) = 177º (clearly not what we're looking for)
So it seems the direction in which to head is about 27.7º east of north
As in the start, this direction can be expresed in (x,y) coordinates by using
x = r sin? ; y = r cos?
so it becomes; v = (23.2x + 44.3y) km/h.