Velocity Vectors of a Boat Relative to the Shore and River

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SUMMARY

The discussion focuses on calculating the velocity of a motorboat traveling due east at 15 mi/h relative to the shore, while the river flows southwest at 4 mi/h. The total velocity of the boat is determined through vector summation, resulting in a formula that combines the contributions of both the boat and the water. The final velocity vector is expressed as v = <-4 cos(45deg) + 15, -4sin(45 deg)>, with the speed calculated as the magnitude of this vector. The direction is determined using arctan, indicating the heading relative to the moving water.

PREREQUISITES
  • Understanding of vector summation
  • Knowledge of trigonometric functions (sine and cosine)
  • Familiarity with the concept of relative velocity
  • Basic skills in solving equations involving magnitudes and directions
NEXT STEPS
  • Study the double subscript method for relative velocities
  • Learn about vector addition in physics, particularly in fluid dynamics
  • Explore the application of trigonometric functions in navigation problems
  • Investigate the effects of current on boat navigation using real-world examples
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and vector analysis, as well as anyone involved in navigation and maritime operations.

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Homework Statement


The water in the river moves southwest (45 degrees south of west) at 4 mi/h.
If a motorboat is traveling due east at 15 mi/h relative to the shore, determine the speed of the boat and its heading relative to the moving water.

Homework Equations


Vector summation

The Attempt at a Solution


My attempt:
The water making the boat move 4 mi/h, 45 degrees south of west means the the water s contribution to the boat s total velocity is v_1 = <-4 cos(45deg), -4sin(45 deg)>.
Furthermore, the boat's motor's contribution to the boat's velocity is v_2 = <15,0>.
Therefore the total velocity, by vector addition, is v = v_1 + v_2 = <-4 cos(45deg), -4sin(45 deg)> + <15,0> = <-4 cos(45deg)+15, -4sin(45 deg)+0> = <-4 cos(45deg)+15, -4sin(45 deg)>.
Then, speed is the magnitude of the velocity vector. So, ||v|| = sqrt([-4 cos(45deg)+15]^2 + [-4sin(45 deg)]^2), and its direction is arctan( [-4 cos(45deg)+15] / [-4sin(45 deg)]) south of east.

Questions about my attempt:
Is the velocity I find relative to the water or shore?

I suspect it's relative to the shore and that the velocity relative to the "moving water" is simply 15 mph i^ (i hat). Am I right? If so, could someone please let me know? If not, could someone please elaborate?

Any input would be greatly appreciated!
 
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