A ship leaves its home port expecting to travel to a port 600 km due south. Before it moves even 1 km, a severe storm blows it 120 km due east. How far is the ship from its destination?
In what direction must it travel to reach its destination?
---° (south of west)
2007-11-02 17:59:51 · 5 answers · asked by matt w 1 in Science & Mathematics ➔ Physics
Do you have a textbook? Do you have the basic math and geometry prerequisites for the physics class? Use them.
2007-11-02 18:29:56 · answer #1 · answered by Frank N 7 · 1⤊ 0⤋
We can assume that the ship was 600 km away from the port before the storm had blown. The storm displaces it to 120 km due east.
We can form a right angled triangle of height 600 km and base 120 km joining the port, the original position of the ship and the new position. The hypotnuse of this triangle will give the distance of the ship from the port.
d = square root of( 120^2 + 600^2) km
= 620 km approximately.
2007-11-03 01:10:25 · answer #2 · answered by raj 2 · 0⤊ 0⤋
direction of ship was south but the storm blew it to east that is there is an angle of 90 between them.
This can be reffered to as a right angled triangle.The displacement between the destination and current position is equal to the length of the hypotenuse.
By pythagoras theorum:
Hypotenuse^2=sum of the squares of the sides.
displacement^2=(600^2+120^2)
displacement^2=374400
displacement=661.882341631Km.
so,it must travel in south west direction.
2007-11-03 02:18:00 · answer #3 · answered by srejangoyal 1 · 1⤊ 0⤋
I'm starting to doubt that your teacher didn't teach all of these things. Did you miss a day of class or something?
Use the Pythagorean Theorem and the law of cosines.
2007-11-03 01:04:00 · answer #4 · answered by Ben 3 · 1⤊ 0⤋
sowth-west is the direction
use pythogras therom....
2007-11-03 01:13:30 · answer #5 · answered by M B 2 · 0⤊ 0⤋