The entrance to a harbour is a channel of length a which runs between two sandbanks a distance b apart. The banks and the channel can be assumed to be rectangular. On this particular day, there is a current of constant speed v flowing from one sandbank to the other. A yacht travelling at a constant speed relative to the water of u wants to enter the harbour.

Find the least value of u needed for the yacht to safely enter the harbour.

Someone please help me solve this and/or provide an answer, I'm struggling.

Find the least value of u needed for the yacht to safely enter the harbour.

Someone please help me solve this and/or provide an answer, I'm struggling.

Original post by student273

The entrance to a harbour is a channel of length a which runs between two sandbanks a distance b apart. The banks and the channel can be assumed to be rectangular. On this particular day, there is a current of constant speed v flowing from one sandbank to the other. A yacht travelling at a constant speed relative to the water of u wants to enter the harbour.

Find the least value of u needed for the yacht to safely enter the harbour.

Someone please help me solve this and/or provide an answer, I'm struggling.

Find the least value of u needed for the yacht to safely enter the harbour.

Someone please help me solve this and/or provide an answer, I'm struggling.

This is the classic riverboat vector problem with the usual flow of the river and boat direction transposed:

The x direction is also the length of the channel.

The y direction is the channel width and river flow direction.

The boat therefore has a y velocity component given by the river current.

So the problem is to derive an expression for the speed and direction of the boat relative to the river flow, in order to transit the length of the channel in the time it takes to cross the channel.

HINT: redraw the above diagram to comply with the labels defined by the question you posted.

(edited 6 years ago)

Original post by uberteknik

This is the classic riverboat vector problem with the usual flow of the river and boat direction transposed:

The x direction is also the length of the channel.

The y direction is the channel width and river flow direction.

The boat therefore has a y velocity component given by the river current.

So the problem is to derive an expression for the speed and direction of the boat relative to the river flow, in order to transit the length of the channel in the time it takes to cross the channel.

HINT: redraw the above diagram to comply with the labels defined by the question you posted.

The x direction is also the length of the channel.

The y direction is the channel width and river flow direction.

The boat therefore has a y velocity component given by the river current.

So the problem is to derive an expression for the speed and direction of the boat relative to the river flow, in order to transit the length of the channel in the time it takes to cross the channel.

HINT: redraw the above diagram to comply with the labels defined by the question you posted.

Thanks, but I'm still really confused as to how I give my answer (U) in terms of V

Original post by student273

Thanks, but I'm still really confused as to how I give my answer (U) in terms of V

The boat will drift at the same velocity as the current Vms

$V_{drift} = \frac{b}{t_{drift}}$

rearranging for t

$t_{drift} = \frac{b}{V_{drift}}$................eq 1

For safe passage, the boat must traverse the length of the channel (a) metres before the time t

$U_{boat} = \frac{a}{t_{drift}}$................eq 2

substituting eq 1

$U_{boat} = \frac{a}{(\frac{b}{V_{drift}})}$

$U_{boat} = (\frac{a}{b})V_{drift}$

removing the subscripts and complying with the minimum time requirement:

$U > \frac{aV}{b}$

(edited 6 years ago)

Original post by uberteknik

The boat will drift at the same velocity as the current Vms^{-1} between the sandbanks. The distance between the sandbanks is given as (b) metres and takes time t_{drift} seconds to cross from one side to the other.

$V_{drift} = \frac{b}{t_{drift}}$

rearranging for t_{drift}:

$t_{drift} = \frac{b}{V_{drift}}$................eq 1

For safe passage, the boat must traverse the length of the channel (a) metres before the time t_{drift} elapses.

$U_{boat} = \frac{a}{t_{drift}}$................eq 2

substituting eq 1

$U_{boat} = \frac{a}{(\frac{b}{V_{drift}})}$

$U_{boat} = (\frac{a}{b})V_{drift}$

removing the subscripts and complying with the minimum time requirement:

$U > \frac{aV}{b}$

$V_{drift} = \frac{b}{t_{drift}}$

rearranging for t

$t_{drift} = \frac{b}{V_{drift}}$................eq 1

For safe passage, the boat must traverse the length of the channel (a) metres before the time t

$U_{boat} = \frac{a}{t_{drift}}$................eq 2

substituting eq 1

$U_{boat} = \frac{a}{(\frac{b}{V_{drift}})}$

$U_{boat} = (\frac{a}{b})V_{drift}$

removing the subscripts and complying with the minimum time requirement:

$U > \frac{aV}{b}$

Hi there, I completely understand the answer that you have given, but when I enter it apparently the boat is not necessarily travelling perpendicular to V therefore it cannot be U=av/b

Original post by student273

Hi there, I completely understand the answer that you have given, but when I enter it apparently the boat is not necessarily travelling perpendicular to V therefore it cannot be U=av/b

Working and diagram

I realise that this is quite some time after requested, but hopefully this is useful to others:

Starting from eq 2 from @uberteknik:

a can be expressed in terms of u as

a = ut / (cos(θ))

so u = (a * cos(θ))/t

Now, we can solve for cos(θ)

Answer

I realise that this is quite some time after requested, but hopefully this is useful to others:

Starting from eq 2 from @uberteknik:

a can be expressed in terms of u as

a = ut / (cos(θ))

so u = (a * cos(θ))/t

Now, we can solve for cos(θ)

•

cos(θ) = b / (b^2+a^2)^(1/2) eq 3 Where (b^2+a^2)^(1/2) is the minimum distance the boat must travel, so gives minimum u

•

Now if we substitute eq 3and eq 1 back into eq 2

•

We get:

•

u = ab*v / b*(b^2+a^2)^(1/2)

•

Which cancels to give:

•

u = av / (b^2+a^2)^(1/2)

•

Which Isaac Physics had told me is the right answer

Answer

(edited 11 months ago)

- What are these a levels and their courses like?
- Is the physics in an Engineering degree much harder than A Level physics?
- How to Ace Physics
- What is COMPOS?
- Screwed for Cambridge application?
- I want to do a comp sci degree but no physics a level pls
- 4 weeks till paper 1 - how to improve at AQA physics?
- Deciding on Physical Sciences
- A seemingly unachievable goal
- is it a good choice to take biology physics further maths without chem?
- Does further maths help with maths entrance exams?
- Entry requirements for Physics and Astrophysics degree
- Mental health nursing or Social care
- I crashed higher geography and im worried I wont pass
- Studying mechanical engineering as an economic student
- Advice for getting A*s in A levels
- Erection Problems?
- Wanting to study A levels as an adult - help!
- How useful is doing a course in physics (A-Level) for studying medicine?
- GCSE Physics Study Group 2023-2024

Latest

Trending