Respuesta :
Answer:
The angle at which the boat must head is [tex]- 22.47^{\circ}[/tex]
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:
[tex]f(x) = 3sin(\frac{\pi x}{40})[/tex] (2)
Now, to determine the angel at which the boat must head:
The velocity of the engine of the boat:
v = [tex]u'cos\theta\hat{i} + u'sin\theta\hat{j}[/tex]
v = [tex]5cos\theta \hat{i} + 5sin\theta\hat{j}[/tex]
The abscissa of the boat at time t:
v = [tex]5cos\theta t\hat{i}[/tex]
Now, from above and eqn(1) , we can write:
[tex]f(5cos\theta t) = 3sin(\frac{\pi \times 5cos\theta t}{40})[/tex]
Now, boat's velocity at time t:
v = [tex]5cos\theta \hat{i} + (5sin\theta + 3sin(\frac{\pi \times 5cos\theta t}{40})\hat{j}[/tex]
In order to obtain the position of the boat, we integrate both the sides, we get:
r = [tex]5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j}[/tex] + C (3)
Now, at r = 0:
0 = [tex]5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j}[/tex] + C
C = [tex]\frac{24}{\pi cos\theta}\hat{j}[/tex]
Now, from eqn (3)
r = [tex]5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40} + \frac{24}{\pi cos\theta})\hat{j}[/tex] (4)
the baot will reach the point at y = 0 and x = 40
Now,
40 = [tex]5cos\theta t[/tex]
[tex]t = \frac{8}{cos\theta}[/tex]
Substituting the above value of 't' in eqn (4):
r = [tex]5sin\theta \frac{8}{cos\theta}\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta \frac{8}{cos\theta}}{40} + \frac{24}{\pi cos\theta})\hat{j}[/tex]
We get:
[tex]48 + 40\pi sin\theta = 0[/tex]
[tex]\theta = sin^{- 1}(\frac{- 48}{40\pi}) = - 22.47^{\circ}[/tex]
