At noon, ship A is 100 Kilometers due east of ship B, Ship A is sailing west at 12 k/h, and ship B is sailing S10degrees west at 10k/h. At what time will the ships be nearest each other and what time will the distance be?
(not a right triangle)?
Not a right triangle?
You know angle, or SAS.
you want the other distance (S)
I think I would start with the law of cosines:
c^2=a^2+b^2 -2ab CosC
Ok. So c^2=100^2+b^2-2(100)b Cos100 ?
c^2=100^2-x+(-12t)^2-2(100-x)(-12t) Cos100 or maybe this?
Still trying to figure this out.
Am i going in the right direction?
c2=(12t)^2+(10t)^2-2(12t)(10t)Cos100
To solve this problem, we can use the concept of relative velocity and distance formula in a coordinate system.
Step 1: Set up a coordinate system
Let's set up a coordinate system with Ship B's initial position as the origin (0, 0), and the east direction as the positive x-axis, and the north direction as the positive y-axis.
Step 2: Convert the sailing directions and speeds
Since Ship A is sailing west, its velocity vector can be written as (-12, 0) km/h. Ship B is sailing S10 degrees west. We need to convert this velocity vector into the coordinate system we defined. To do this, we can use trigonometry.
The angle S10 degrees west is equivalent to 100 - 10 = 90 degrees since west is perpendicular to north. So, Ship B's velocity vector can be written as (cos(90), sin(90))*10 = (0, 10) km/h.
Step 3: Set up the equation for the distance between the ships
The distance between two points (x1, y1) and (x2, y2) can be calculated using the distance formula:
distance = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, we want to find the time when the ships are nearest, i.e., when the distance between them is minimum. So, we can square this distance equation to eliminate the square root:
distance^2 = (x2 - x1)^2 + (y2 - y1)^2
Step 4: Find the time when the ships are nearest
To find the time when the ships are nearest, we can find the minimum value of the squared distance equation.
Using the time variable t, we can express the positions of Ship A and Ship B over time as follows:
Position of Ship A: (100 - 12t, 0)
Position of Ship B: (0, 10t)
Substituting these positions into the squared distance equation, we get:
distance^2 = ((100 - 12t) - 0)^2 + (0 - 10t)^2
distance^2 = (100 - 12t)^2 + (10t)^2
To find the time when the ships are nearest, we need to minimize this equation. We can achieve this by taking the derivative of this equation with respect to t, setting it equal to zero, and solving for t.
Step 5: Solve for the time and distance
Taking the derivative of the distance^2 equation with respect to t, we get:
2(100 - 12t)(-12) + 2(10t)(10) = 0
-2400 + 288t + 200t = 0
488t = 2400
t ≈ 4.92 hours (approximately)
This means that after approximately 4.92 hours (or 4 hours and 55 minutes), the ships will be the nearest to each other.
To find the distance at this time, we can substitute t = 4.92 into the distance equation:
distance = √((100 - 12(4.92))^2 + (0 - 10(4.92))^2)
distance ≈ 56.68 kilometers (approximately)
So, after approximately 4.92 hours, the ships will be nearest at a distance of approximately 56.68 kilometers.