May I have some help with this question? I'm not sure how to solve it.
A boat leaves Kingston and heads due east for 25 km. At the same time, a second boat travels in a direction 30° south of east from Kingston for 15 km. How far apart are the boats at this moment when they reach their destinations?
Thank you!
After you make your sketch and let x km be the distance between them, you will see that
x^2 = 25^2 + 15^2 - 2(25)(15)cos30°
....
Thanks!
All angles are measured CW from +y-axis.
KA = 25km[90o].
AK = 25km[90+180].
KB = 15km[120o].
AB = AK+KB = 25[270o]+15[120o]
AB = (25*sin270+15*sin120)+(25*cos270+15*cos120)i
AB = -12-7.5i = 14.2km[58o]W. of S. = 14.2km[238o] = distance between the boats.
Of course, I can help you with that!
To solve this question, we can use the concept of vector addition. We will break down the motions of each boat into their respective components and then add those components to find the distance between the boats.
Let's start with the first boat that travels due east for 25 km. Since it is going due east, its motion can be represented by the vector (25, 0), where the first number represents the displacement in the east direction and the second number represents the displacement in the north direction (which is zero in this case).
Now, let's consider the second boat that travels in a direction 30° south of east from Kingston for 15 km. We will first calculate its displacement in the east direction and then its displacement in the north direction.
The displacement in the east direction can be calculated as follows:
east displacement = 15 km * cos(30°)
east displacement = 15 km * √(3)/2
east displacement ≈ 12.99 km
Similarly, the displacement in the north direction can be calculated as follows:
north displacement = 15 km * sin(30°)
north displacement = 15 km * 1/2
north displacement = 7.5 km
So, the motion of the second boat can be represented by the vector (12.99, -7.5), where the positive east displacement is represented by the first number and the negative north displacement is represented by the second number.
Now, we will add the two vectors to get the total displacement between the boats:
total displacement = (25, 0) + (12.99, -7.5)
total displacement = (25 + 12.99, 0 + (-7.5))
total displacement = (37.99, -7.5)
To find the distance between the boats at this moment, we can use the Pythagorean theorem. The distance between the boats is the magnitude of the total displacement vector:
distance = √(37.99^2 + (-7.5)^2)
distance = √(1443.0201 + 56.25)
distance ≈ √(1499.2701)
distance ≈ 38.73 km
So, the boats are approximately 38.73 km apart when they reach their destinations.
I hope that helps! Let me know if you have any further questions.