word problem-distance between ships?

Question

two ships leave the same harbor at the same time. the first ship heads north at 20mph and the second ship heads weat at 15mph. what is the expression that gives the distance "d" between the ships after "t" hours?

i'm a bit confused. do i need to use the distance formula and the "speed=distance/time" formula to solve this?

thanks for the help. i really appreciate it.

edit-change weat to west. sorry.

Answer 1

A diagram would make things clearer, but the important point to take away is that the ships are traveling perpendicular to each other. They are essentially creating a triangle, with each ship's path being one leg and the hypotenuse being the distance between them. This is a right triangle, since the legs are perpendicular. We know from Pythagoras that we can find the hypotenuse if we know the lengths of the two legs. So all we need to find are two equations,each representing the distance the ships have traveled as a function of time.

You already cited this equation in your question: distance = velocity*time.
For ship one, traveling at 20mph, Distance=20t
For ship two, Distance=15t

So, the total distance between them as a function of time will equal, by Pythagoras:
D = Sqrt((Distance Ship 1)^2 + (Distance Ship 2)^2)
D = Sqrt((20t)^2 + (15t)^2)
D = Sqrt(400t^2 + 255t^2)
D = Sqrt(625t^2)
D = 25t

Answer 2

You actually have to use the Pythagorean theorem in this one, because the ships are traveling at right angles to each other. The distance between the two ships will be the square root of the sum of the squares of the distances each ship has taken. In time t, the first ship goes 20t miles north and the second ship goes 15t miles west, so

D = sqrt (400t^2 + 225t^2) = sqrt (625 t^2) = 25t

Which means D=25t is the distance between the ships. I hope that makes sense. If you draw a picture, you should see this as a right triangle that just keeps expanding as the ships head away from where they started.

Answer 3

The ships are following 2 sides of a right triangle. The distance between them is found from a2 + b2 = c2 or
c = Sqrt(a2+b2)

side 1 15 (a2 = 225)
side 2 20 (b2 = 400)

400mi+225mi = 625mi

side c (aka distance between ships after 1 hour) = 25mi

The ships are moving apart at a steady 25 miles per hour.
After two hours 50 miles appart, 75, 100..... ect..

Answer 4

The easiest way to do it is break it down like a triangle and use Pythagorean Theorem to determine distance.

x^2 + y^2 = z^2

The ship heading north can be represented by 20x (20 mph times x time traveled)

The west ship can be represented by 15y (15 mph times y time traveled)

And the distance between them is the square root of z.

Then you get (20x)^2 + (15y)^2 = sqrt z

You need to have x and y be the same time...

Example - if time = 2 hours then...

(20*2)^2 + (15*2)^2 = sqrt 2500

Square root of 2500 is 50...That means after 2 hours, the ships are 50 miles apart...

Answer 5

distance travelled in the northern direction in t hs=20t miles and the distance travelled by the other ship in the Western direction in t hrs will be 15t miles

[(20t)^2+(15t)^2]^1/2 will be the formula for the distance in t hrs
Pythagoras theorem gives the hypotenuse

Answer 6

if the distance traveled from the first ship is d1 and from the second ship is d2, then

d1 = v1*t, d1 = 20t
d2 = v2*t, d2 = 15t

The distance d between the two ships is from Pythagorean theorem:

d = sqrt(d1^2 + d2^2)

so:

d = sqrt(400*t^2 + 225*t^2)
d = sqrt(625*t^2)
d = 25t

Answer 7

25*t

20:15 = 4:3...it's a right trangle. So the distance increases at 25 mph.

Answer 8

Are the ships on water?

Answer 9

hummmm. School must be back in session. Good Luck.