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 straight stretch of river flowing north, with parallel banks 40 m apart. If the maximum water speed is 3 m/s, we can use the sine function, f(x) = 3 sin(πx/40), as a basic model for the rate of water flow x units from the west bank. Suppose a boater would like to pilot the boat to land at the point B on the east bank directly opposite point A. If the boat maintains a constant heading and a constant speed of 5 m/s, determine the angle at which the boat should head. (Round your answer to one decimal place.)

Q = 20.91 degrees.

Explanation:

Given:

- The velocity of boat in x direction v_x = 5 m/s

- The velocity of boat in y direction f(x) = 3sin(pi*x/40)

- The bank are x distance apart x = 40m

Find:

Determine the angle at which the boat should head.

Solution:

- We will first see that how much down the river will the boat be thrown at if boat doesn't move at an angle initially.

- The distance covered in x direction is:

                                x = v_x*t

                                x = 5*t

- The velocity of water is in y-direction is:

                                v_y = 3*sin(pi*x/40)

- Substitute the distance x traveled found above into v_y:

                               v_y = 3*sin(pi*5*t/40)

                               v_y = 3*sin(pi*t/8)  

- What we have is the function of velocity with respect to time t. we know that velocity is the derivative of displacement. Hence,

                               y = integral ( v_y ) .dt

                               y = integral ( 3*sin(pi*t/8) ) .dt          

- Perform integration:

                               y = -3*8/pi*cos(pi*t/8) + C      

                               y = -24/pi *cos(pi*t/8) + C        

- We know at t = 0, y = 0. Evaluate C:

                               0 = -24/pi *1  + C        

                               C = 24/pi

Hence,

                                y = 24/pi *( 1 - cos(pi*t/8) )  

- Back substitute t = x / 5, we have:

                                y = 24/pi *( 1 - cos(pi*x/40) )

- Now compute y(40):

                                y = 24/pi * ( 1 - cos(pi) )

                                y = 24*2 / pi = 48 / pi

- Now we compute the angle:

                               Q = arctan ( y / x )

                               Q = arctan ( 48/40*pi )

                               Q = 20.91 degrees.

- Hence, the boat has to traverse 20.91 degrees towards the flow or river to reach on the other side of the bank just ahead.

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

higher potential

explanation:

because it is in a high speed so it must move into a of high potential