2 ships leave port at the same time. The first ship heads due north at 5miles per hour while the second...?

Question

2 ships leave port at the same time. The first ship heads due north at 5miles per hour while the second heads due west at 3 miles per hour. Let d be the distance between teh ships in miles and let t be the time in hours since they left port. Find d as a function of t

Answer 1

It's very unfortunate, but your question can only be answered assuming the Earth is flat, but without that assumption I can't see how you can solve it (e.g. "going west" only means a straight line if you're on the equator).

Distance of first ship from the port - 5*t

Distance of second ship from the port = 3*t

d = sqrt [(5*t)^2 + (3*t)^2]  = t * sqrt(34)

Answer 2

The two ships will traverse the sides of a right triangle, with the hypotenuse being the distance between them.

Distance of the north ship is 5t

Distance of the west ship is 3t

The separation between the ships can be found using the Pythagorean Theorem

A^2 + B^2 = C^2

So...

d^2 = (5t)^2 + (3t)^2

or

d^2 = 25t^2 + 9t^2

d^2 = 34t^2

so

d = sqrt(34)t

Note that you don't need to worry about +- issues, as time and distances must all be positive.

Answer 3

distance = speed x time

In t hours the ship heading north would have traveled 5t miles and the ship heading west would have traveled 3t miles.

Using the Pythogorean Theorem, d^2=(5t)^2+(3t)^2

d^2=25t^2+9t^2

d^2=34t^2

d=sqrt(34) t miles in t hours.

Answer 4

d(t)=sqrt(25t^2+9t^2) or t*sqrt(34)

Answer 5

d(t)=sqrt(5t^2+3t^2)