# River-boat problem (Relative velocity)

• baldbrain
In summary, a man wants to cross a river in the shortest time possible, taking into account the river's width of 120 m, his rowing speed of 3 m/s in still water, his maximum walking speed of 1 m/s, and the river's flow velocity of 4 m/s. To find the shortest time, the man must minimize the total time, which is a combination of the time taken to cross the river and the time needed to compensate for downstream drift. This can be achieved by finding the optimal angle at which to row the boat, as it will determine the drift and thus the additional time needed to walk the boat either upstream or downstream.
baldbrain

## Homework Statement

A man wishes to cross a river of width 120 m by a motorboat. His rowing speed in still water (or relative to water) is 3 m/s and his maximum walking speed is 1 m/s. The river flows with a velocity of 4 m/s.
(a) Find the path which he should take to reach the point directly opposite to his starting point in the shortest time.
(b) Find the time required to reach the destination.

2. Homework Equations

Let ω be the width of the river and x be the drifting of the boat.
Let v denote the velocity vectors and v denote their respective magnitudes.
vr = absolute velocity of river
vbr = velocity of boatman relative to the river or velocity of boatman in still water
vb = absolute velocity of boatman
Hence, vbr = vb - vr
vb = vbr + vr
Time taken to cross the river t = ω/vbr-y = ω/vbrcos θ
Drift x = vb-xt = (vr - vbrsin θ)ω/vbrcos θ

## The Attempt at a Solution

Now, in order to reach the point directly opposite to his starting point, resultant velocity vb must be perpendicular to the river current.
So, for this the drift x must be 0.
x = 0 gives vr = vbrsin θ
Subsituting, sin θ = 4/3, which is meaningless.
I can't proceed without θ. Also, I don't get why they've given his maximum walking speed.
Am I missing anything?

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You are missing that a 3 m/s velocity cannot have a 4 m/s component so regardless of how he rows he will have to walk back on the other side. This is why you are given the walking speed. The angle ##\theta## is a parameter and you will obtain different results for different angles. You are asked to find the shortest time possible so you need to minimise the time as a fuction of the angle.

So, t = 120/3cos θ = 40/cos θ
x = (4 - 3sin θ)40 sec θ = 160sec θ - 120tan θ
But I can't just differentiate t = 40sec θ. That wouldn't render anything...

Please define your variables properly. If not you are just assuming we know what you are doing, but we will only be guessing. In particular, what is your definition of the time t and what is its relation to your sought time?

Orodruin said:
Please define your variables properly. If not you are just assuming we know what you are doing, but we will only be guessing. In particular, what is your definition of the time t and what is its relation to your sought time?
Of course, it's the time taken to cross the river. We must find the shortest time possible, which, as you said, must be minimised as a function of the angle.
But I still don't get how to apply the walking speed and what to do with the drift equation.

I have the drift of the boat as a function of the angle.

baldbrain said:
Of course, it's the time taken to cross the river. We must find the shortest time possible, which, as you said, must be minimised as a function of the angle.
We are not asked to minimize the time taken to cross the river. We are asked to minimize the time taken to reach a point opposite the starting point.

Since the boat is guaranteed to drift downstream as it is rowed across the river, the man must either walk the boat upstream before launching or after landing on the opposite bank. How far he has to walk it will depend on the angle at which he rows.

jbriggs444 said:
How far he has to walk it will depend on the angle at which he rows.
So we take x to be the drift of the man and then find the time (of course as a function of the angle)?
That means time of walking t| = x/vwalking
∴ t| = (160sec θ - 120tan θ)/1
total time = t + t| = 200sec θ - 120tan θ
And, then you minimize it? Does that sound right?

baldbrain said:
So we take x to be the drift of the man and then find the time (of course as a function of the angle)?
That means time of walking t| = x/vwalking
∴ t| = (160sec θ - 120tan θ)/1
total time = t + t| = 200sec θ - 120tan θ
And, then you minimize it? Does that sound right?
WIthout having carefully gone over the arithmetic, yes, that sounds right.

It has the right qualitative feel. It has the 200 sec θ which embodies the elapsed time crossing the river plus the extra time needed to compensate for downstream drift. It has the -120 tan θ which embodies the limited advantage one can get by aiming upstream to reduce the drift velocity.

jbriggs444 said:
WIthout having carefully gone over the arithmetic, yes, that sounds right.

It has the right qualitative feel. It has the 200 sec θ which embodies the elapsed time crossing the river plus the extra time needed to compensate for downstream drift. It has the -120 tan θ which embodies the limited advantage one can get by aiming upstream to reduce the drift velocity.
That would yield θ = sin-1 (3/5) and t = 160 s.
Thanks

## What is the "River-boat problem" and why is it important in relative velocity?

The "River-boat problem" is a classic physics problem that involves determining the relative velocity of an object in a moving river. It is important in relative velocity because it demonstrates how the motion of the water affects the motion of the boat, and how to calculate the net velocity of the boat relative to the shore.

## How do you calculate the relative velocity of a boat in a river?

To calculate the relative velocity of a boat in a river, you first need to determine the velocity of the boat in still water. Then, you need to factor in the velocity of the river, which is typically given in the problem. Finally, use vector addition to find the net velocity of the boat relative to the shore.

## What are the common misconceptions when solving the "River-boat problem"?

One common misconception is that the velocity of the boat in still water and the velocity of the river will cancel each other out, resulting in a net velocity of zero. However, this is incorrect as the two velocities are in different directions and must be added using vector addition. Another misconception is assuming that the boat is always traveling at a constant speed, when in reality it may be accelerating or decelerating.

## How does the direction of the river affect the relative velocity of the boat?

The direction of the river can greatly affect the relative velocity of the boat. If the river is flowing in the same direction as the boat, the relative velocity will be greater, resulting in a faster net velocity. If the river is flowing in the opposite direction, the relative velocity will be smaller, resulting in a slower net velocity. The angle at which the river flows can also impact the direction and magnitude of the relative velocity.

## How is the "River-boat problem" applied in real-life situations?

The "River-boat problem" is commonly used in navigation and transportation industries to determine the most efficient route for boats and ships to travel. It is also used in hydrology to study the flow of rivers and how it impacts the environment. Additionally, it can be applied in sports such as rowing and kayaking to understand the effects of currents on the boats and athletes.

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