math prob.?

Question

Suppose a person walks one kilometer east, then one kilometer northeast, then another kilometer east. find the distance in kilometers, between the person's intitial and final locations.

Answer 1

This question can not be answered. It depends on initial location of the person on the earth. When the person is close to the pole and walks 1 km to the east (or west) he can pass his inital position multiple times since he walks in a circle around the pole.

Answer 2

OK... here's the process

First lets rearrange the order of the steps - lets say they walk 1km East, then anothe 1km east then 1km NE, giving us the path below:

_ _ /

As mentioned earlier the obvious solution is to use pythagorous, so we want to draw a right angled triangle. Let the longest side be the length from the start to the finish, and the other two sides will look something like this

_ _ _ |

The height of the triangle is from the NE journey, where 1^2=a^2 + b^2
Luckily a = b so we get 1=2a^2 or a=1/root2

and the longer side is made up of 2 km plus another a, or 1/root2

Finally we use pythagorous again to find the distance d

d^2 =(1/root2)^2 + (2+ 1/root2)^2
d^2 =0.5 + (9+4root2)/2
d^2 = 5+2root2
d = 2.7979

Answer 3

Just over 2 kilometers

Answer 4

Let us say we start at point O and go 1km east to A then we go NE for 1km that is at 45degrees to OA and away from OA. Let the point where 1km ends be B then he goes 1km east that is like parallel to OA starting at B to C the final point. We need to find out the distance OC the actual distance between first point and final point.

This looks something like

B C
` __
__/
O A

Now joining the ends to form a rectangle we get something like this. We need to get the distance of diagonal OC

__ __
|__/ __|

The side will be 1km + 1/(2^1/2) + 1km and the height will be
1/(2^1/2) let this be equal to a

by pythogorean theorem OC^2 = (2km + a)^2 + a^2

= 4 + 4a + a^2 + a^2
= 4 + 4a + 2*a^2
= 4 + 4a + 1
= 5 + 4a
OC = (5 + 4a)^1/2
= (5 + 4/(2^1/2))^1/2
= 2.7979 km ~ 2.8km

Answer 5

Start at (0, 0).
Walk 1 km east to (1, 0).
Walk 1 km northeast to (1 + [√(2)] / 2, [√(2)] / 2).
Walk 1 km east to (2 + [√(2)] / 2, [√(2)] / 2).

d² = {2 + [√(2)] / 2}² + {[√(2)] / 2}²
d² = [9 / 2 + 2√(2)] + (1 / 2)
d² = 5 + 2√(2)

d = √[5 + 2√(2)], or about 2.7979326519318134 kilometers.

Answer 6

3 kilometers

Answer 7

1 km due E and then 1 km due northeast and then 1km due East again and so joining the starting point and the finishing point the distance=
[(1+2^-1/2)+1]^2+(2^-1/2)^2]^1/2 km=[{1+2(2)^1/2}^2/4+1/2]^1/2km

Answer 8

2.5 Kilometers? Its just a guess.

Answer 9

It's a simple trig problem. Easiest way I find is to draw it out and us pythagoren's theorm. When you draw it out it should look like a strechted out, backwards Z. connect the 2 farthest points and you get 2 equal triangles. Fint the hypothenuse on one and double it and that's your answer. I got 2 times the square root of 1.25 which is about 2.36 km.

Answer 10

Will want an empty field Fill 5 litre Fill 3 litre from 5 litre........leaves 2 litres in 5 litre field Empty into spare field or drink Repeat and to that end get or drink 4 litres or you ought to empty the three litre and then pass the two litres from 5 litre to now empty 3litre fill the 5 litre and fill the three litre from it. this could take a million litre from the 5 litre leaving 4 litres