# Question #5d5a2

##### 1 Answer

#### Answer:

Here's how you can solve this problem.

#### Explanation:

*Water of a river, that is 1000m wide and has parallel banks, flows towards north in a constant velocity of* *A man rows a boat from a place in a bank with a constant velocity of* *to the east, relative to the water.*

*What is the boat's velocity relative to earth?**When the boat reaches the next bank, what is the distant it has flowed down the river?**What is the time taken for the man to cross the river?*

*If the man wants to get to the point perpendicular to the point he starts rowing, with all the other vectors same as above*

*What is the direction he has to row the boat?**Then, what is the boat's velocity relative to the earth?**What is the time taken for the man to reach the other bank of the river?*

So, you know that the river is **1000 m** wide. The boat moves with a *constant speed* of *constant speed* of

If you take the north direction to be the

#v^2 = v_x^2 + v_y^2#

The magnitude of the velocity of the boat **relative to the bank** of the river will be equal to

#v = sqrt(v_x^2 + v_y^2)#

#v = sqrt(3^2 + 2^2)"ms"^(-1) = color(green)("3.61 ms"^(-1))#

Now, the boat will move at an angle of

#tan(theta) = v_y/v_x implies theta = arctan(v_y/v_x)#

#theta = arctan((2color(red)(cancel(color(black)("ms"^(-1)))))/(3color(red)(cancel(color(black)("ms"^(-1)))))) = color(green)(33.7""^@)#

The velocity of the boat will thus be *north of due east*.

You can find the distance the boat travelled *down river* by taking into account the fact that the **width** of the river is covered with a speed of

Likewise, the **distance travelled downstream** will be covered with a speed of

This means that you can find the **time** needed to cross the river by using

#underbrace(d_"x")_(color(blue)("width of river")) = v_x * t#

#t = d_x/v_x = (1000color(red)(cancel(color(black)("m"))))/(3color(red)(cancel(color(black)("m")))"s"^(-1)) = "333.3 s"#

Since the time needed to cross the river is *the same* for both directions, you can say that

#underbrace(d_y)_(color(blue)("distance downstream")) = v_y * t#

#d_y = 2"m"color(red)(cancel(color(black)("s"^(-1)))) * 333.3color(red)(cancel(color(black)("s"))) = color(green)("667 m")#

Now, for the second part of the problem, you need to find the direction the boat must take in order to reach the other bank in a point **directly across** its starting point.

All you really need to do here is *flip the triangle vertically* formed with the width of the river and the distance travelled downstream.

Since all vectors remain unchanged, it follows that the boat must start rowing at an angle of

#theta_2 = -33.7""^@#

The *minus sign* symbolizes the fact that this angle is measured *below* the

The magnitude of the velocity of the boat will be the same

#v = sqrt(v_x^2 + v_y^2)#

#v = sqrt(3^2 + 2^2)"ms"^(-1) = "3.61 ms"^(-1)#

The velocity of the boat will be *south of due east*.

Of course, the boat will reach the other bank in the same period of time,

Here's what I mean by *flipping the triangle*. The boat starts crossing the river, and

If all vectors are kept the same, all you have to do is

This time, the boat will start moving at an angle *vertical component*, which is the speed of the river current*, and the *horizontal component*, which is the *speed of the boat*, are the same.