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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.
1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
d1 =
2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, as in part (a)), how far downstream is she swept before reaching the opposite bank?
d2 =
3) For the conditions of part (b), how long does it take for her to reach the opposite bank?
t =

Answer :

Answer:

1) [tex]\Delta s=1000\ ft[/tex]

2)  [tex]\Delta s'=998.11\ ft.s^{-1}[/tex]

3) [tex]t\approx125\ s[/tex]

[tex]t'\approx463.733\ s[/tex]

Explanation:

Given:

width of river, [tex]w=500\ ft[/tex]

speed of stream with respect to the ground, [tex]v_s=8\ ft.s^{-1}[/tex]

speed of the swimmer with respect to water, [tex]v=4\ ft.s^{-1}[/tex]

Now the resultant of the two velocities perpendicular to each other:

[tex]v_r=\sqrt{v^2+v_s^2}[/tex]

[tex]v_r=\sqrt{4^2+8^2}[/tex]

[tex]v_r=8.9442\ ft.s^{-1}[/tex]

Now the angle of the resultant velocity form the vertical:

[tex]\tan\beta=\frac{v_s}{v}[/tex]

[tex]\tan\beta=\frac{8}{4}[/tex]

[tex]\beta=63.43^{\circ}[/tex]

  • Now the distance swam by the swimmer in this direction be d.

so,

[tex]d.\cos\beta=w[/tex]

[tex]d\times \cos\ 63.43=500[/tex]

[tex]d=1118.034\ ft[/tex]

Now the distance swept downward:

[tex]\Delta s=\sqrt{d^2-w^2}[/tex]

[tex]\Delta s=\sqrt{1118.034^2-500^2}[/tex]

[tex]\Delta s=1000\ ft[/tex]

2)

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

[tex]v'=v.\cos37[/tex]

[tex]v'=4\times \cos37[/tex]

[tex]v'=3.1945\ ft.s^{-1}[/tex]

Now the net effective speed of stream sweeping the swimmer:

[tex]v_n=v_s-v'[/tex]

[tex]v_n=8-3.1945[/tex]

[tex]v_n=4.8055\ ft.s^{-1}[/tex]

The  component of swimmer's velocity heading directly towards the opposite bank:

[tex]v'_r=v.\sin37[/tex]

[tex]v'_r=4\sin37[/tex]

[tex]v'_r=2.4073\ ft.s^{-1}[/tex]

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

[tex]\tan\phi=\frac{v_n}{v'_r}[/tex]

[tex]\tan\phi=\frac{4.8055}{2.4073}[/tex]

[tex]\phi=63.39^{\circ}[/tex]

  • Now let the distance swam in this direction be d'.

[tex]d'\times \cos\phi=w[/tex]

[tex]d'=\frac{500}{\cos63.39}[/tex]

[tex]d'=1116.344\ ft[/tex]

Now the distance swept downstream:

[tex]\Delta s'=\sqrt{d'^2-w^2}[/tex]

[tex]\Delta s'=\sqrt{1116.344^2-500^2}[/tex]

[tex]\Delta s'=998.11\ ft.s^{-1}[/tex]

3)

Time taken in crossing the rive in case 1:

[tex]t=\frac{d}{v_r}[/tex]

[tex]t=\frac{1118.034}{8.9442}[/tex]

[tex]t\approx125\ s[/tex]

Time taken in crossing the rive in case 2:

[tex]t'=\frac{d'}{v'_r}[/tex]

[tex]t'=\frac{1116.344}{2.4073}[/tex]

[tex]t'\approx463.733\ s[/tex]

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