A jogger runs directly east for 3.0 km, then turns and goes Northwest for 5.0 km.?

Question

He then travels directly outh for 2.0 km. How far and in what direction is he from the starting point?

Thank you very much!

Answer 1

It depends where he starts from.  If he starts need the equator, then he ends up about 2 km north of were he starts.  But if he starts near the north pole, he could be 6.4 km from where he started.

Answer 2

Assuming that the directions are fixed meaning every time you say south or east it always means the same and not relative to his current direction.

I think he will end up 2 Km north of his starting position.

Explanation:

His movements are tracing a right triangle which have the measurements of 3-4-5Km  with the 3 Km line being the base, 5Km line being the hypotenuse and the 4Km line being the vertical line. And since he completes only 3 of the last 4 Kms needed to complete the triangle, he ends up 2Kms north of the starting point.

_ |.\

2  |...\  5

_  |.... \

2  |____\

_       3

So, he's somewhere on the vertical line.

I know u probably think I'm really jobless to be drawing this but.. what'd i do? Just finished my engineering. I am jobless. :-)

Footnote: The only calculations are: 3^2=4^2=5^2

=> 9+16=25

This verifies that it does form a right triangle, which proves that i was right about his path leading him back to where he started. But since he only completed 2 Kms, it means he's 2Km short of the starting point.

Answer 3

1.62 km & roughly northwest of starting point.  

Ok, the diagrams are messy 'cos it's hard to draw here, but  hope you can understand; point x =starting point, point f = final point

step1) draw a right angle triangle (Isosceles), you get this:

go east then turn northwest gives 45 degrees

go northwest then turn south gives 45 degrees

______________|\ <----45 degrees

______________| __\

_______length L  | ____\     5km

______________| ______\

______________| ________\              

90 degrees  ------>|___________\ <---- 45 degrees

________________length L

step 2) L= 5 cos(45) = 3.53km  

step 3)

3.53km -2 km = 1.53km

3.53km - 3km = 0.53km  

---> gives new triangle:

______ f

______|\

__1.53 |   \

______|___\X  

______0.53

dist between x & f = the diagonal length

= square root (1.53^2 + 0.53^2) = 1.62km

from the diagram you can tell that f is northwest of x

Answer 4

C2=A2+B2

C2=25+9

C=34^

C=5.8km