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You want to cross a river that is 30 m wide and flows to the East at 3.0 m/s . You can swim at a speed of 5.0 m/s (relative to the water). If you start from the South side of the river and want to go straight across, at what angle should you point yourself

Sagot :

jacob193

Answer:

Approximately [tex]\text{N$37^{\circ}$W}[/tex] (approximately [tex]37^{\circ}[/tex] west of north.)

Explanation:

The velocity of the swimmer relative to the water is given to be [tex]v(\text{swimmer})= 5.0\; {\rm m\cdot s^{-1}}[/tex]. Decompose this velocity into two components:

  • parallel to the flow of the river: [tex]v_{\parallel}(\text{swimmer})[/tex];
  • perpendicular to the flow of the river: [tex]v_{\perp}(\text{swimmer})[/tex].

Decompose the velocity of the current [tex]v(\text{current}) = 3.0\; {\rm m\cdot s^{-1}}[/tex] along the same set of directions.

  • parallel to the flow of the river: [tex]v_{\parallel}(\text{current}) = 3.0\; {\rm m\cdot s^{-1}}[/tex] to the east;
  • perpendicular to the flow of the river: [tex]v_{\perp}(\text{current}) = 0\; {\rm m\cdot s^{-1}}[/tex] (Note that the current in the river flows along the river, not toward the banks of the river.)

The velocity of the swimmer relative to the bank would be the vector sum [tex](v(\text{swimmer}) + v(\text{current}))[/tex]. Decompose this vector sum in the same directions:

  • parallel to the flow of the river: [tex]v_{\parallel}(\text{swimmer}) + v_{\parallel}(\text{current})[/tex];
  • perpendicular to the flow of the river: [tex]v_{\perp}(\text{swimmer}) + 0\; {\rm m\cdot s^{-1}} = v_{\perp}(\text{swimmer})[/tex].

Swimming straight across requires that the swimmer travel towards the banks but not along the banks. In other words, the velocity of the swimmer relative to the bank should be [tex]0[/tex] in the direction of the river: [tex]v_{\parallel}(\text{swimmer}) + v_{\parallel}(\text{current}) = 0[/tex].

Rearrange to obtain:

[tex]v_{\parallel}(\text{swimmer}) = -v_{\parallel}(\text{current})[/tex].

Thus, [tex]v_{\parallel}(\text{swimmer})[/tex] should be the exact opposite of [tex]v_{\parallel}(\text{current})[/tex].

Since [tex]v_{\parallel}(\text{current}) = 3.0\; {\rm m\cdot s^{-1}}[/tex] to the east, [tex]v_{\parallel}(\text{swimmer})\![/tex] should be [tex]3.0\; {\rm m\cdot s^{-1}\![/tex] to the west.

Refer to the diagram attached: [tex]v(\text{swimmer})[/tex] and its two components [tex]v_{\parallel}(\text{swimmer})[/tex] and [tex]v_{\perp}(\text{swimmer})[/tex] are the three sides of a right triangle, with[tex]v(\text{swimmer})= 5.0\; {\rm m\cdot s^{-1}}[/tex] as the hypotenuse.

Since the swimmer is travelling to the north, [tex]v_{\perp}(\text{swimmer})[/tex] would point northward.

The side [tex]v_{\parallel}(\text{swimmer})[/tex] would be opposite to the angle between the hypotenuse [tex]v(\text{swimmer})= 5.0\; {\rm m\cdot s^{-1}}[/tex] and north ([tex]v_{\perp}(\text{swimmer})[/tex],) Thus, the value of that angle would be:

[tex]\begin{aligned}& \arcsin\left(\frac{\text{opposite}}{\text{hypotenuse}}\right) \\ \\ =\; & \arcsin\left(\frac{3\; {\rm m\cdot s^{-1}}}{5\; {\rm m\cdot s^{-1}}}\right) \\ \approx \; & 37^{\circ}\end{aligned}[/tex].

Since [tex]v_{\parallel}(\text{swimmer})[/tex] points to the west, the direction of [tex]v(\text{swimmer})[/tex] would be approximately [tex]37^{\circ}[/tex] west of north, [tex]\text{N$37^{\circ}$W}[/tex].

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