SOLUTION: A boy standing on the shore of a lake 1 mile wide wants to go to the store 3 miles down the shore on the opposite side of the lake. If he swims at 2 miles per hour, and walks at 4

Algebra ->  Formulas -> SOLUTION: A boy standing on the shore of a lake 1 mile wide wants to go to the store 3 miles down the shore on the opposite side of the lake. If he swims at 2 miles per hour, and walks at 4       Log On


   



Question 292469: A boy standing on the shore of a lake 1 mile wide wants to go to the store 3 miles down the shore on the opposite side of the lake. If he swims at 2 miles per hour, and walks at 4 miles per hour, is it quicker for him to swim directly across the lake and then walk to the store, or is it quicker for him to swim directly to the store? Prove this mathematically.
Found 2 solutions by richwmiller, Alan3354:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
walk 3 and swim 1
4*w=3
w=3/4
2*s=1
s=2/4
s+w=5/4
swim diagonal
1^2+3^2=d^2
1+9=d^2
10=d^2
sqrt(10)=d
3.16=d
2*s=3.16
s=1.58 hours
Faster and safer to walk first.
.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A boy standing on the shore of a lake 1 mile wide wants to go to the store 3 miles down the shore on the opposite side of the lake. If he swims at 2 miles per hour, and walks at 4 miles per hour, is it quicker for him to swim directly across the lake and then walk to the store, or is it quicker for him to swim directly to the store? Prove this mathematically.
----------------
If he swims 1 mile and walks 3:
1/2 + 3/4 = 1.25 hours
-----------------------
If he swims directly to the store, the distance is
sqrt%283%5E2+%2B+1%29+=+sqrt%2810%29
--> ~ 1.58 hours
-----------------
If he swims at an angle of 30 degs toward the store and walks the rest:
--> ~ 1.183 hours, the minimum time