Two boats leave the same port at the same time with one boat traveling north at 15 knots per hour.?

Question

The other boat is traveling west at 12 knots per hour.
How fast is the distance between the two boats changing after 2 hours?

Answer 1

The second answer got it right but he didn't show his work. Here's a solid mathematical proof to answer to your question using a bit of calculus.

You have the equation which describes the situation.

X^2 + Y^2 = Z^2

Take the derivative in respect to T. This is basically saying find the rate at which X Y and Z change in respect to time.

2X (dx/dt) + 2Y (dy/dt) = 2Z (dz/dt)

Simplify
X (dx/dt) + Y (dy/dt) = Z (dz/dt)

We can calculate X and Y and Z (2 hours out to sea at their respective speeds and Z is calculated using the first formula)

(30) (dx/dt) + (24) (dy/dt) = (1476^.5) (dz/dt)

We know what dx/dt and dy/dt are (that's the speed given in the question) we want to find dz/dt.

(30)(15) + (24)(12) = (1476^.5)(dz/dt)
450 + 288 = (1476^.5)(dz/dt)
738/ (1476^.5) = dz/dt

19.20937 = dz/dt

Wohoo! And then I realized that since the velocities are constant, the dz/dt should be constant too... Could have done it like using basic Pythagorean...

So a bit of mental masturbation but what they heck!

You can do it the way the guy beneath me did it but MAKE SURE YOU MULTIPLY THE NUMBERS CORRECTLY!!!!

He get's the same answer as I did but he didn't divide by 2 like he said he did.

Answer 2

Let
2 hr = the time of movement of both boats
15 knots/hr = rate of boat A (due north)
12 knots/hr = rate of boat B (due west)
30 knots north = displacement (distance with direction) of boat A
24 knots west = displacement of boat B
d = distance between both boats.

since north and west form a right triangle, you can use the pythagorean theorem to solve for d.
d² = (30 knots)² + (24 knots)²
d² = 900 knots² + 576 knots²
d² = 1476 knots²
d = 6sqrt41 knots

Since
V = d/t

and we know that
d = 6sqrt41 knots
t = 2 hr
Therefore,

V = (6 sqrt 41 knots)/2 hr
V = 3 sqrt 41 knots/hr
V = 38.4187 knots/hr
Therefore, the distance between the two boats change at the rate of 38.4187 knots/hr.

^_^


One more thing:
A knot is a nautical mile^_^

^_^

Answer 3

d is the distance between the two vehicles, a is the distance from the N vehicle to the port, b the same for the V vehicle.
d^2=a^2+b^2
a=15*t
b=12^t
d=sqrt(225*t^2+144*t^2)=t*sqrt(369)
so the distance is a linear function of time in this case
after two hours: d=2h * sqrt(369)
but you have to modify the measurement units to obtain a right result: velocity in meters/second, time in seconds and you will get the distance in meters
I don't know what a knot is.

Answer 4

3 knots. the distance keeps chaning 15-12=3 knots.

Btw, a knot is a nautic mile per hour. So the speed is just in knots, not "knots per hour".

Answer 5

V = (225 + 144)^(1/2)

V =369^(1/2)
V = 19.20 knots/hour in N-W direction

Answer 6

it's root over (2*12)^2 + (2*15)^2 = 38,42
( a^2 + b^2 = c^2)
(after 2 hours, so 2*12 and 2*15)

Answer 7

The rate of change of distance between 2 boats will remain constant. not sure.........

Answer 8

it's root over369
coz 15^2+12^2 will give you aparting speed

Answer 9

need more information With only the info given, the other boat can travel as fast or slow as it pleases.