Look for speed

Question

I set it up as (2(x+2))²+(2x)²=20² but I'm having trouble solving for speed

 

Two fishing boats depart a harbor at the same time, one traveling east, the other south. The eastbound boat travels at a speed 2 mi/h faster than the southbound boat. After 2 h the boats are 20 mi apart. Find the speed of the southbound boat.

Answer 1

R is the speed of southbound boat.

R+2 is the speed of the eastbound boat.

 

The distance between them, per pythagorean theorem, is  square root of [ (Rt)^2 + ((R+2)t)^2 ]

 

  (2R)^2 + (2(R+2))^2 = 20^2 <---- YES, this is the equation you have

 

 4R^2 + 4(R+2)^2 = 400

 

4R^2 + 4( R^2 + 4r + 4) = 400

 

 

4R^2 + 4r^2 + 16r + 16 = 400

 

8r^2 + 16r - 384 = 0

 

r^2 + 2r - 48 = 0

 

(r   +  8)(r -  6 ) = 0

 

r+8 = 0 ---> r = -8  ---> rejected; speed cannot be negative or else he's going backwards

 

r - 6 = 0 ---> r = 6

 

THe southbound boat is going 6 mph.

The eastbound boat is going 8 mph.

 

In 2 hours they have gone 12 and 16 miles respectively.

 

Answer 2

4 (x² + 4x + 4) + 4x² = 400 =>

 

(x² + 4x + 4) + x² = 100 =>

 

2x² + 4x + 4 = 100  =>

 

x² + 2x + 2 = 50  =>

 

x² + 2x - 48 = 0   =>

 

(x + 8) (x _ 6) = 0 => x = -8 , x= 6 , -8 has no meaning here therefore x = 6 mil/hr .

 

Answer 3

You are correct. Your "x" is the rate of the slow boat. Just continue with the solution!