Question

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 1

1)
`\Delta s=1000\ ft`

2)
`\Delta s'=998.11\ ft.s^(-1)`

3)
`t\approx125\ s`

`t'\approx463.733\ s`

Step-by-step explanation:

Given:

width of river,
`w=500\ ft`

speed of stream with respect to the ground,
`v_s=8\ ft.s^(-1)`

speed of the swimmer with respect to water,
`v=4\ ft.s^(-1)`

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

`v_r=√(v^2+v_s^2)`

`v_r=√(4^2+8^2)`

`v_r=8.9442\ ft.s^(-1)`

Now the angle of the resultant velocity form the vertical:

`\tan\beta=(v_s)/(v)`

`\tan\beta=(8)/(4)`

`\beta=63.43^(\circ)`

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

so,

`d.\cos\beta=w`

`d* \cos\ 63.43=500`

`d=1118.034\ ft`

Now the distance swept downward:

`\Delta s=√(d^2-w^2)`

`\Delta s=√(1118.034^2-500^2)`

`\Delta s=1000\ ft`

2)

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

`v'=v.\cos37`

`v'=4* \cos37`

`v'=3.1945\ ft.s^(-1)`

Now the net effective speed of stream sweeping the swimmer:

`v_n=v_s-v'`

`v_n=8-3.1945`

`v_n=4.8055\ ft.s^(-1)`

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

`v'_r=v.\sin37`

`v'_r=4\sin37`

`v'_r=2.4073\ ft.s^(-1)`

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

`\tan\phi=(v_n)/(v'_r)`

`\tan\phi=(4.8055)/(2.4073)`

`\phi=63.39^(\circ)`

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

`d'* \cos\phi=w`

`d'=(500)/(\cos63.39)`

`d'=1116.344\ ft`

Now the distance swept downstream:

`\Delta s'=√(d'^2-w^2)`

`\Delta s'=√(1116.344^2-500^2)`

`\Delta s'=998.11\ ft.s^(-1)`

3)

Time taken in crossing the rive in case 1:

`t=(d)/(v_r)`

`t=(1118.034)/(8.9442)`

`t\approx125\ s`

Time taken in crossing the rive in case 2:

`t'=(d')/(v'_r)`

`t'=(1116.344)/(2.4073)`

`t'\approx463.733\ s`