An RCMP patrol boat left Long Beach and sailed west for 2 hours at 15 km/h, then North for 2 h at 20 km/h, then east for 1 h at 10 km/h. How far was the patrol boat from Long beach, to the nearest tenth of a kilometre?
So, how would you do this question? What are the steps to solving it?
2007-10-06 05:11:34 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Best to see this by a diagram:
>east 10km
^north 40 km
therefore the boat is 20km west and 40km north of original position.
therefore using pythagorum theory
distance = square root of ((20^2)+(40^2))
= square root of 2000
= 44.72km from Long Beach
2007-10-06 05:23:53 · answer #1 · answered by Anonymous · 0⤊ 0⤋
by placing these simple directions onto a page labled north, west, east, and south, you are practicing vector geometry. The mathematical equivalent would be to use an x y plane, but you should still form a right triangle with a 20 kilometer leg and a 40 kilometer leg. Lable these legs a, b, then call the hypotenuse c. From the pythagorean theorum in trigomometry, a squared plus b squared equals c squared, you should get 44.72 kilometers from long beach.
2007-10-06 12:43:42 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Assume Long Beach is at (0,0)
Then boat is at (-30,0) after going West.
The boat will be at (-30, 40) after North leg.
The boat will be at ( -20,40) after East leg
So we want to find distance from (0,0) to (-20,40).
Use distance formula to get :
d =sqrt(40^2+20^2) = sqrt(2000) = 44.7 km
2007-10-06 12:25:24 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
__________Present position
! 10km
!
!
!40 km
!
!
!
!
!____30km.__________X Long island
The net distance travelled is the diagonal for the sides 20km and 40 km. Right?That is approximately sq.root of
( 400 + 1600 ) = 14.14 kms. O.K.?
2007-10-06 12:30:50 · answer #4 · answered by Pandian p.c. 3 · 0⤊ 0⤋