A ship is being steered due east. A current is flowing from north to south, so that the actual velocity of the ship is 12 km/h on a bearing of 120 degree . Find the speed of the current and still water speed of the ship
To find the speed of the current and the still water speed of the ship, we can use vector addition.
Let's break down the given information:
1. The ship is moving due East. This means the ship's velocity in still water is entirely in the East direction.
2. The actual velocity of the ship is 12 km/h on a bearing of 120 degrees. The "bearing" refers to the angle made with the North direction.
Now, let's solve for the speed of the current and the still water speed of the ship:
1. Start by drawing a vector diagram. Draw the North direction pointing upward, the ship's velocity vector (in still water) pointing due East, and the ship's actual velocity vector (including the effect of the current) at an angle of 120 degrees with the North direction. Label the ship's velocity in still water as "v" and the current velocity as "c".
2. Now, since the ship's actual velocity is the vector sum of the ship's speed in still water and the current's speed, use vector addition. Draw the vector from the end of the ship's velocity in still water to the end of the ship's actual velocity (which represents the current's velocity). The resulting vector is the ship's actual velocity. Label this vector as "v + c".
3. Use the information given in the problem to create a triangle. The magnitude of the ship's actual velocity ("v + c") is given as 12 km/h, and the angle between the ship's actual velocity and the ship's velocity in still water is 120 degrees.
4. Now, split the triangle into two separate right-angled triangles. The triangle with a right angle, the side opposite to the 120-degree angle, and the hypotenuse representing the ship's actual velocity is the resultant velocity triangle.
5. Use trigonometry to solve for the magnitude of the current's velocity ("c"). Since we know the magnitude of the ship's actual velocity ("v + c") and the angle between "v + c" and "v," we can use the cosine rule:
(v + c)^2 = v^2 + c^2 - 2vc * cos(120)
Simplify this equation, keeping in mind that "v" represents the speed of the ship in still water:
12^2 = v^2 + c^2 + 2vc * 0.5
144 = v^2 + c^2 + vc
6. You also know the relationship between the speed of the current ("c") and the ship's speed in still water ("v") from the given information in the problem. When the ship is steering due East, the angle between the ship's velocity in still water and the current is 90 degrees.
7. Use trigonometry again, this time using the sine rule, to solve for the magnitude of the current's velocity ("c"):
sin(90) = c / v
1 = c / v
c = v
8. Substitute the value of "c" from step 7 into the equation from step 6:
144 = v^2 + v^2 + v^2
Simplify this equation:
144 = 3v^2
48 = v^2
9. Take the square root of both sides to solve for the ship's speed in still water ("v"):
sqrt(48) = v
v ≈ 6.93 km/h (rounded to two decimal places)
10. Since the speed of the current ("c") is equal to the ship's speed in still water ("v") from step 7, the speed of the current is also approximately 6.93 km/h.
Therefore, the speed of the current and the still water speed of the ship are both approximately 6.93 km/h.
Course made good = 120 deg (math teachers do not do navigation and always misuse the word bearing. Moreover ships do knots which are nautical miles per hour, not km/hr, but anyway) which is east + 30 deg south
East component of course made good = 12 cos 30 = 10.4 km/h
South component of course made good = 12 sin 30 = 6 km/h
So
speeds
Vs = speed of ship east
Vc = speed of current south
East speeds
Vs + 0 = 10.4 = ship speed
South speeds
0 + Vc = 6 = current speed