(b) The current in a river flows at 1.51 ms-1 relative to the shore. A swimmer with a velocity of 3.76 ms-1 relative to the water wishes to swim directly across the river. At what angle to the shore should the swimmer head in order to arrive directly opposite on the other bank? What is the swimmer’s velocity relative to an observer on the shore.

Question

(b)	The current in a river flows at 1.51 ms-1 relative to the shore. A swimmer with a velocity of 3.76 ms-1 relative to the water wishes to swim directly across the river. At what angle to the shore should the swimmer head in order to arrive directly opposite on the other bank? What is the swimmer’s velocity relative to an observer on the shore.

i used pythagoras and got 4.05 for the velocity and inverse tan of 1.5/3.76 and got 21.7 degrees upstream ????

Steve · Answer

The swimmer's direction should be the hypotenuse of the diagram. Then, the current makes the resultant velocity perpendicular to the bank.

The angle is thus upstream at an angle of arccos(1.5/3.76)=66.5° relative to the bank.

That makes the speed relative to the shore √(3.76^2 - 1.5^2)=3.45m/s directly across the river. (Assuming the observer stands where the swimmer took off.)

Anonymous · Answer

inverse cos (1.51/3.76) = 66.3 square root (3.76^2/1.51^2) = 4.05

Anonymous · Answer

inverse cos (1.51/3.76) = 66.3 square root (3.76^2 - 1.51^2) = 3.44 sorry slight adjusment remember you're no finding the longest side