At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 20 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.)
It still says NOT CORRECT.
To solve this problem, we can use the Pythagorean theorem to find the distance between the ships at any given time. Let's break down the problem step-by-step:
Step 1: Define the variables
Let's assign variables to the quantities mentioned in the problem:
- DA: The distance ship A has traveled (in nautical miles)
- DB: The distance ship B has traveled (in nautical miles)
- t: Time elapsed since noon (in hours)
- x: The distance between the ships at time t (in nautical miles)
Step 2: Derive equations
Based on the information given, we can derive equations for DA, DB, and x using the following data:
- Ship A is sailing west at 17 knots, so DA = 17t
- Ship B is sailing north at 20 knots, so DB = 20t
- Using the Pythagorean theorem, we can find x at any given time:
- x^2 = (DB)^2 + (DA - 10)^2
Step 3: Find the value of x at 4 PM
To find the value of x at 4 PM, we need to substitute t = 4 into the equation for x:
- x^2 = (20 * 4)^2 + (17 * 4 - 10)^2
- Simplifying further:
- x^2 = 400 + (68 - 10)^2
- x^2 = 400 + 58^2
- x^2 = 400 + 3364
- x^2 = 3764
- x ≈ √3764
- x ≈ 61.3 nautical miles
Step 4: Calculate the rate of change of x
To find how fast the distance between the ships is changing at 4 PM, we can differentiate the equation for x with respect to time:
- 2x(dx/dt) = 2(DB)(d(DB)/dt) + 2(DA - 10)(d(DA)/dt)
We already have the values for DB, DA, dx/dt, d(DB)/dt, and d(DA)/dt:
- x = 61.3 nautical miles
- DB = 20t (20 nautical miles per hour)
- DA = 17t (17 nautical miles per hour)
- t = 4 hours
Substituting these values into the equation and solving for dx/dt:
- 2(61.3)(dx/dt) = 2(20 * 4)(20) + 2(17 * 4 - 10)(17)
- 2(61.3)(dx/dt) = 2(80)(20) + 2(58)(17)
- (61.3)(dx/dt) = (80)(20) + (58)(17)
- (61.3)(dx/dt) = 1600 + 986
- (61.3)(dx/dt) = 2586
- dx/dt ≈ 42.2 knots
Therefore, the distance between the ships is changing at a rate of approximately 42.2 knots at 4 PM.
To find the rate at which the distance between the ships is changing, we can use the concept of relative velocity. The relative velocity of ship A with respect to ship B can be calculated by subtracting the velocity of ship B from the velocity of ship A.
Given:
Velocity of ship A, vA = 17 knots (westward)
Velocity of ship B, vB = 20 knots (northward)
At noon, ship A is 10 nautical miles due west of ship B. To determine the distance between the ships at that time, we can use the Pythagorean theorem:
Distance^2 = (10 nautical miles)^2 + (0 nautical miles)^2
Distance = 10 nautical miles
Now, to find the rate at which the distance between the ships is changing at 4 PM, we need to consider the time elapsed from noon to 4 PM, which is 4 hours.
Using the concept of relative velocity, we can calculate the rate of change of distance as follows:
Rate of change of distance = relative velocity = √[(velocity of ship A)^2 + (velocity of ship B)^2]
Applying the values:
Rate of change of distance = √[(17 knots)^2 + (20 knots)^2]
Rate of change of distance ≈ √[289 + 400] knots ≈ √689 knots ≈ 26.24 knots
Therefore, the distance between the ships is changing at a rate of approximately 26.24 knots at 4 PM.
let t hours be some time past 12:00 noon
After t hours, ship B has gone 20t knots
and ship A has gone 17t knots
let the distance between them be D
I see a right-angled triangle and
D^2 = (20t)^2 + (17t+10)^2
D^2 = 689t^2 + 340t + 100
2D(dD/dt) = 1378t + 340
dD/dt = (689t + 170)/D
at 4:00 pm, t=4 and
D^2 = 12484
D = 111.7318
so at t=4
dD/dt = (689(4) + 170)/111.7318
= 26.19 knots