you have to use Pythagorean to solve.
After 3 hours, it has traveled 45 km north (rate of 15km/hr)
After 5 hours, it has traveled 85 km west (rate of 17 km/hr)
Now you have a right triangle, now solve for the hypotenuse (the distance from the ship to the port)
Pythagorean is a^2+b^2=c^2 (we are solving for 'c')
solving for c, you get sqrt(45^2+85^2) = sqrt(2025+7225)
Thus, distance is 96.18 km
Hope this helps.
2006-10-13 06:43:46
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answer #1
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answered by JSAM 5
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This is a simple geometry problem (assuming no currents, a constant speed and an infinitely sharp turn after three hours):
First draw a triangle with a vertical line 15*3 (45km) long AND then a horizontal line 17*5 (85km) long
Then find the length of the hypotenuse connecting the start point from the end (the third leg of the triangle) to discover how far the ship is from port.
Since it is a right triangle, the hyp = sqrt(45^2+85^2) = 96.18km from port after 8 hours.
2006-10-13 06:55:08
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answer #2
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answered by figurehead 2
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If you plot the ship's course on a piece of paper taking the North to be the top of the page, south to be the bottom, and east to the right and west to the left, you will see that the ship has made a right angled triangle.
The distance traveled is the hypotenuse of the triangle.
After 3 hours the ship is 45 km away from the port going straight north. 5 hours after changing directions to west, the ship is 85 km away from the northernmoist point.
So here you have a right angled triangle with length of both the arms of the triangle given. Just calculate the hypotenuse using pythagoras theorem and there you have it.
Hint: Direction changes from North-West, West-South, and South-East, and East-North and vice versa are at right angles.
2006-10-13 06:53:05
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answer #3
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answered by Anonymous
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The best way to solve this would be as follows:
First figure how far it went to the north which is done by multiply the rate times the time which comes out to 15 km/hr times 3 hours which equals 45 km. Next do the same with the distance west and you get 85 km. The using the following formula you can find the distance form start to finish: C squared = a squared + b squared or c squared would be 9250 taking the square root of that you get approximately 96.18 km from the start
2006-10-13 06:50:04
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answer #4
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answered by rkvideo 2
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This question has no answer :
It depends how far the port is from the northpole.
suppose it is 50 km from the north pole.
after 3 hours it is 5 km from the northpole
if it goes west for 5 hours at 17 kkm/h it travels 85 kms this is aprooximately 6 rounds around the north pole and the ship is at the same place when it started to go to the west.
Tell this to your teacher
2006-10-13 06:55:08
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answer #5
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answered by gjmb1960 7
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Gimb has got the right idea, but expressed its detail unclearly, as he/she assumes that a line of latitude 5 kms south of the north pole is equivalent to a circle of 5 kms radius (if it were, its 2 x pi x r circumference would be 31,4159 kms long and two-and a bit circumnavigations of the pole not six would be all you would manage on a 85 kms journey westwards)
However Gimb is absolutely correct that there is no clear unique answer. And that the teacher hasn't realised that!
The answer could be as much as the approx 96.18 kms most people's Pythagorean triangle suggests, or it could be as little as 45 kilometres as I argue below, or it could be any number inbetween, all depending on where your starting point is, vis-a-vis the North Pole.
HOW DO I CONTEND THE DISTANCE FROM PORT COULD BE AS LITLE AS 45 KILOMETRES?
Assuming that the polar ice cap is not in the way i.e. it is all clear water in a circumpolar navigation of the globe (i.e. we are on a mathematical ideal planet or on earth after global warming has melted the polar ice caps), if the ship starts 45 kms south of the point where one circumpolar journey around a line of latititude would be 85 kilometres long, then travelling west for 85 kms will bring you back to the point where the northern leg of the jouney ended and the ship is then only 45 hilometres from its starting point,
There are an infinity of such solutions of course, i.e. starting from a point 45 kms south of the line of latitude where a 42.5 kms westwards circumpolar journey would bring you back to the point where you finished sailing north, you would make two complete circuits of that line of latitide and the ship will then be only 45 kilometres from the starting point,
Similarly start 45 kms south of a line of latitude which is 85/3 kms in circumference, where 3 circuits of the line of latitude are completed by sailing west, and the ship will then be only 45 kilometres from the starting point, etc.
Depending on the size of the planet you are on, there may perhaps be one or more solutions at the south pole, too.
Of course we are assuming there is a land mass with a port on its northern coast at the methematically precise place from which we want the journey to start.
No Pythagorean triangle or calculation will be necessary in all such cases.
The Flat Earth Society will tell you that the ship will fall off the edge of the globe into the abyss, if you attempt circumnavigation pf the globe but you can safely ignore them, I think. Ferdinand Magellan did!
2006-10-13 08:27:06
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answer #6
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answered by Anonymous
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North for 3 hours at 15km/h is 45 kilometers North
East for 5 hours at 17km/h is 85 kilometers East
Using the pythagorean theorem: a²+b²=c²
45²=2025
85²=7225
2025+7225=9250
√9250=96.17692031 kilometers
2006-10-13 06:47:35
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answer #7
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answered by Anonymous
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First of all change the sign an the 3 to a single sign. The equation would read -45=x-3+50. 1) -45=x-3+50 (collect like terms...the -3 and +50) 2) -45=x+47 3) -47 -47 Answer -92 REMEMBER THE NEGATIVE
2016-05-21 23:10:06
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answer #8
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answered by ? 4
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North dist.travelled in 3 hrs@15kmph=45km
West dist. travelled in 5hrs@17kmph=85km
We observe that the ship has traversed a
right angled triangle. The dist. we want
is the hypotenuse of the triangle.
Dist.from port=sqrt[45^2+85^2]
=5sqrt[9^2+17^2]
=5*sqrt380
=5*19.49
=97.46 km
2006-10-13 06:58:52
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answer #9
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answered by openpsychy 6
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distance in the northerndirection=45 km
distance in the western direction=85 km
distance across
using c^2=a^2+b^2
c^2=45^2+85^2
=2025+7225
c=square root of 9250
=96.18
distance from the port=96.18km
2006-10-13 06:43:45
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answer #10
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answered by raj 7
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