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Difficulty: MediumScalars and Vectors

A boat moves due north across a river at a speed of 4.0 m s−14.0\text{ m s}^{-1} relative to the water. If the river current flows due east at a speed of 3.0 m s−13.0\text{ m s}^{-1}, what is the magnitude of the resultant velocity of the boat relative to the riverbank?

  1. A
    1.0 m s−11.0\text{ m s}^{-1}
  2. 5.0 m s−15.0\text{ m s}^{-1}Answer
  3. C
    7.0 m s−17.0\text{ m s}^{-1}
  4. D
    25.0 m s−125.0\text{ m s}^{-1}

Answer

The magnitude of the resultant velocity of the boat relative to the riverbank is 5.0 m s−15.0\text{ m s}^{-1}.
The resultant velocity magnitude is found by applying vector addition for perpendicular components: vresultant=(4.0)2+(3.0)2=25.0=5.0 m s−1v_{resultant} = \sqrt{(4.0)^2 + (3.0)^2} = \sqrt{25.0} = 5.0\text{ m s}^{-1}.

Step-by-Step Solution

1
Identify the given velocity components and their directions
Northward velocity component vy=4.0 m s−1v_y = 4.0\text{ m s}^{-1}, Eastward velocity component vx=3.0 m s−1v_x = 3.0\text{ m s}^{-1}.
The motion of the boat and the movement of the river current act in perpendicular directions (90∘90^\circ to each other).
2
Calculate the resultant magnitude using vector addition (Pythagorean theorem)
v=vx2+vy2=(3.0)2+(4.0)2=9.0+16.0=25.0=5.0 m s−1v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0\text{ m s}^{-1}.
For two orthogonal vectors, the magnitude of the vector sum equals the hypotenuse of the right triangle formed by the vector components.

Key Concept

Vector Addition of Perpendicular Quantities
Estimated Time:1m 0s
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Topics

Scalars and Vectors

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