If a ship leaves port at 9:00 a.m. and sails due south for 3 hours at 14 knots, then turns N 60° E for another 2 hours, how far from port is the ship?
A. 14 nm
B. 17.75 nm
C. 21.5 nm
D. 22.6 nm
PLEASE HELP!
looks like a rather straight-forward case of the cosine law.
distance^2 = 42^2 + 28^2 - 2(42)(28)cos60
= 1372
distance = 37.04 nm
or
vector of first leg = (42cos270, 42sin270)=(0, -42)
vector for 2nd leg = (28cos30,28sin30) = (24.249, 14)
adding them to get (24.249 , -28)
magnitude = √(24.249^2 + (-28)^2) = √1372 = 37.04 nm
None of your answers match this.
My answer was obtained in 2 totally different methods
Wel, I'm not sure, I have two other questions with practically the same things, please take a look!
If a ship leaves port at 9:00 a.m. and sails due east for 3 hours at 10 knots, then turns N 60° E for another hour, how far from port is the ship?
A. 35 nm
B. 39 nm
C. 43 nm
D. 47 nm
__________________________________
If a ship leaves port at 9:00 a.m. and sails due south for 3 hours at 14 knots, then turns N 60° E for another 2 hours, how far from port is the ship?
A. 14 nm
B. 17.75 nm
C. 21.5 nm
D. 22.6 nm
_____________________________
Thank you; you'tr a life saver!
If a ship leaves port at 9:00 a.m. and sails due north for 3 hours at 12 knots, then turns N 30° E for another hour, how far from port is the ship?
A. 45 nm
B. 47 nm
C. 51 nm
D. 53 nm
This time I get a triangle with sides 30 and 10 and the contained angle between them is 150° , (90+60)
again by the cosine law,
dist^2 = 30^2 + 10^2 - 2(30)(10)cos150
= 1519.62
distance = √1519.62 = appr 38.98 nm
which is choice B
The next question you posted is a repeat of your first
The third:
same setup;
dist^2 = 36^2+ 12^2 - 2(36)(12)cos150
= 2188.25
dist = 46.78 which looks like it is B.
Something fishy about your first one, since I followed exactly the same steps
check for data or answers.
Thank you very much; but these are the only answers! I don't know why, but I'll try to check with my teacher. Again, thanks for the help and showing your work, you're awesome!
Just tried you first question again, using cos30° in my cosine law
and got 22.6 which is D
BUT, according to your wording of N 60° E, the angle between the two paths would be 60, and not 30
So to get your answer it should have been N 30° E
To determine the distance from the port, we can use the concept of vector addition.
First, let's find the distance traveled during the first leg of the journey when the ship sailed due south for 3 hours at 14 knots. Speed is given in knots, and we know that distance = speed × time. Therefore, the distance traveled in the first leg is 14 knots × 3 hours = 42 nautical miles (nm).
Next, let's calculate the distance covered during the second leg of the journey when the ship turned N 60° E for 2 hours. The ship's speed is not provided, but since it is the same ship, we can assume a constant speed throughout the journey. Let's say the speed is 's' knots. Using trigonometry, we can determine the component of the ship's speed in the N 60° E direction.
The N 60° E direction forms a right-angled triangle with the east and north directions. The angle between the N 60° E direction and the east direction is 30 degrees (90 - 60). Therefore, the component of the ship's speed in the N 60° E direction is s × cos(30°).
Since the ship traveled for 2 hours, the distance traveled in the second leg is (s × cos(30°)) knots × 2 hours = 2s × cos(30°) knots.
Now, we have the following information:
- Distance traveled in the first leg: 42 nm
- Distance traveled in the second leg: 2s × cos(30°) nm
To find the distance from the port, we need to find the resultant distance from the starting point, which is the sum of the distances traveled in both legs. Applying the concept of vector addition, we use the Pythagorean theorem:
Resultant distance = √(42^2 + (2s × cos(30°))^2)
Since we want to find the distance from the port, we need to solve for 's'.
Now, we need more information about the ship's speed during the second leg (s). If there is additional information provided or assumed, please mention it.