Question

Water traveling along a straight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at the banks. Consider a long stretch of river flowing north, with parallel banks 40 m apart. If the maximum water speed is 3 m/s, a reasonable model for the water speed of a river x units from the west bank is a sine function:f(x) = 3 sin (πx/40).If a boater would like to cross theriver from a point on the west bank and land at a point exactly opposite with constant headin and a constant speed of 5 m/s, determine the angle at which the boat should head.

Answer 1

The angle at which the boat must head is
`- 22.47^(\circ)`

Solution:

As per the solution:

Distance between the parallel banks, d = 40 m

The maximum speed of water, v' = 3 m/s

constant speed, u' = 5 m/s

Also,

The speed of water of the river at a distance of 'x' units from the west bank is given as a sine function:

`f(x) = 3sin((\pi x)/(40))` (2)

Now, to determine the angel at which the boat must head:

The velocity of the engine of the boat:

v =
`u'cos\theta\hat{i} + u'sin\theta\hat{j}`

v =
`5cos\theta \hat{i} + 5sin\theta\hat{j}`

The abscissa of the boat at time t:

v =
`5cos\theta t\hat{i}`

Now, from above and eqn(1) , we can write:

`f(5cos\theta t) = 3sin((\pi * 5cos\theta t)/(40))`

Now, boat's velocity at time t:

v =
`5cos\theta \hat{i} + (5sin\theta + 3sin((\pi * 5cos\theta t)/(40))\hat{j}`

In order to obtain the position of the boat, we integrate both the sides, we get:

r =
`5sin\theta t\hat{i} + (5sin\theta - (3* 40)/(5\pi cos\theta)cos((\pi * 5cos\theta t)/(40))\hat{j}` + C (3)

Now, at r = 0:

0 =
`5sin\theta t\hat{i} + (5sin\theta - (3* 40)/(5\pi cos\theta)cos((\pi * 5cos\theta t)/(40))\hat{j}` + C

C =
`(24)/(\pi cos\theta)\hat{j}`

Now, from eqn (3)

r =
`5sin\theta t\hat{i} + (5sin\theta - (3* 40)/(5\pi cos\theta)cos((\pi * 5cos\theta t)/(40) + (24)/(\pi cos\theta))\hat{j}` (4)

the baot will reach the point at y = 0 and x = 40

Now,

40 =
`5cos\theta t`

`t = (8)/(cos\theta)`

Substituting the above value of 't' in eqn (4):

r =
`5sin\theta (8)/(cos\theta)\hat{i} + (5sin\theta - (3* 40)/(5\pi cos\theta)cos((\pi * 5cos\theta (8)/(cos\theta))/(40) + (24)/(\pi cos\theta))\hat{j}`

We get:

`48 + 40\pi sin\theta = 0`

`\theta = sin^(- 1)((- 48)/(40\pi)) = - 22.47^(\circ)`