(1 pt) At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 21 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
Relative to ship B's starting point, with t = 0 at noon, the vector position of ship A is
R1 = (-50 -15t) i
and the vector position of ship B is
R2 = 21 t j.
"i" is an east-pointing unit vector and "j" is a north-pointing unit vector.
At 4 PM, t = 4 hours.
The vector that separates R2 and R1 is
R2 - R1 = 21 t j + (50 + 15 t) i.
The rate of change of the separation is
d(R2-R1)/dt = 21 j + 15 i.
Note that is is independent of time.
The magnitude of that vector,
sqrt[(15^2 + 21^2],
is the separation speed that you want
let t hours be some time after noon
(so 4:00 pm is t=4)
so you have a right angled triangle with a vertical of 21t nautical miles and a horizontal of (15t + 50) nautical miles
let the distance between them, or the hypotenuse, be s nautical miles
s^2 = (21t)^2 + (15t+50)^2
2s(ds/dt) = 441t + 30(15t+50)
so when t=4
s^2 = 441(4^2) + 110^2
s = √19156
ds/dt = (441x4 + 30(110))/(2√16156)
= 18.29
so at 4:00 pm the distance between them is changing at 18.29 knots
check my arithmetic.
thanks for helping me but it keeps telling me the answer 18.29 knots is wrong
A tag boat travels at the rate of 15 knots how many hours will i take the boat to travel the distance of 42 nautical miles
To find how fast the distance between the ships is changing at 4 PM, we can use the concept of rate of change, specifically, the derivative. Let's break down the problem and find the solution step by step.
1. We are given that at noon, ship A is 50 nautical miles due west of ship B. This means that at that time, the distance between the ships is 50 nautical miles.
2. Ship A is sailing west at a speed of 15 knots. This means that ship A's x-coordinate (westward distance) changes at a rate of 15 nautical miles per hour.
3. Ship B is sailing north at a speed of 21 knots. This means that ship B's y-coordinate (northward distance) changes at a rate of 21 nautical miles per hour.
4. Since the distance between the ships is changing, let's represent it by a variable, say D(t), where t represents time in hours and D represents the distance between the ships at time t.
5. Using the Pythagorean theorem, we can express the distance D between the ships at any time t as:
D^2 = (x_A - x_B)^2 + (y_A - y_B)^2
where (x_A, y_A) and (x_B, y_B) are the positions of ship A and ship B, respectively.
6. Let's differentiate both sides of the equation with respect to time t:
2D * dD/dt = 2(x_A - x_B) * (dx_A/dt - dx_B/dt) + 2(y_A - y_B) * (dy_A/dt - dy_B/dt)
7. Since we are interested in finding the rate of change of the distance between the ships at 4 PM, we need to find the values of the quantities involved at that time.
8. From noon (t = 0) to 4 PM (t = 4), it has been 4 hours. Since ship A is sailing west at a constant speed of 15 knots, its x-coordinate has changed by 15 * 4 = 60 nautical miles (since speed * time = distance).
9. Ship B is sailing north at a constant speed of 21 knots, and since it has been 4 hours, its y-coordinate has changed by 21 * 4 = 84 nautical miles.
10. Substituting these values into the expression we obtained in step 6, we have:
2D * dD/dt = 2(60 - 0) * (15 - 0) + 2(0 - 84) * (0 - 21)
Simplifying,
2D * dD/dt = 120 * 15 - 168 * (-21)
2D * dD/dt = 1800 + 3528
2D * dD/dt = 5328
11. We know that at noon, the distance between the ships is D = 50 nautical miles. Substituting this value into the equation above, we can solve for dD/dt:
2 * 50 * dD/dt = 5328
100 * dD/dt = 5328
dD/dt = 5328 / 100
dD/dt = 53.28 knots
Therefore, the rate at which the distance between the ships is changing at 4 PM is 53.28 knots.