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Chapter 5: Problem 34

The ratio of the distance carried away by the water current, downstream, in crossing a river, by a person, making same angle with downstream and upstream is \(2: 1\). The ratio of the speed of person to the water current cannot be less than (1) \(1 / 3\) (2) \(4 / 5\) (3) \(2 / 5\) (4) \(4 / 3\)

Short Answer

Expert verified
The minimum ratio is \(\frac{1}{3}\), which is option (1).

Step by step solution

01

Understanding the Problem

We are given a scenario where a person is crossing a river. The ratio of the downstream and upstream distances carried away by water is 2:1. We need to find the minimum possible ratio of the speed of the person (relative to the water) to the speed of the water current itself.
02

Identify Relevant Variables

Let's define the speed of the person as \(v_p\) and the speed of the water current as \(v_w\). The person moves downstream carried by the current, and we know the effect of the current is doubled compared to upstream. Therefore, downstream, the effective speed is \(v_p + v_w\) and upstream is \(v_p - v_w\).
03

Establish Equation with Distance Ratios

The problem gives us that the ratio of distances carried downstream to upstream is \(\frac{2}{1}\). If we assume time taken to cross is the same on either way, the ratio of these effective speeds will give us that ratio:\[\frac{v_p + v_w}{v_p - v_w} = \frac{2}{1}\]
04

Solve for Speed Ratio

From the equation \(\frac{v_p + v_w}{v_p - v_w} = 2\), we solve for \(v_p\) in terms of \(v_w\). Cross-multiplying gives \[v_p + v_w = 2(v_p - v_w)\]Simplifying leads to\[v_p + v_w = 2v_p - 2v_w\]\[3v_w = v_p\]Thus, the ratio \(\frac{v_p}{v_w} = 3\). The inverse gives the minimum ratio as \(\frac{1}{3}\).
05

Conclusion

The ratio of the speeds of the person to the water current cannot be less than \(\frac{1}{3}\), matching option (1) in the choices given.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Upstream and Downstream Motion
When a person or object moves in a river or stream, the water not only flows in its natural course but also affects how the object moves. This is what we call upstream and downstream motion. The object's effective speed changes because of the water's influence, which can either help move it faster or slow it down.

When moving downstream, the person is moving with the current. So, the effective speed of the person becomes the sum of their speed and the speed of the current:
  • Downstream speed: \(v_p + v_w\)
Conversely, when moving upstream, the person is working against the current, so their effective speed is reduced:
  • Upstream speed: \(v_p - v_w\)
Understanding these principles helps explain why the downstream distance is generally greater, especially for longer time periods.
It's crucial to grasp these dynamics because they form the basis of many problems involving river crossing and related scenarios.
Speed of Current
The speed of the current is a crucial factor in river crossing problems and relates to how fast the water is moving in the river. This speed impacts how easily or difficult it is for a person to move through the water, depending on their direction.

Knowing the speed of the current helps in understanding how it affects the total motion:
  • Enhances downstream movement, making it faster and easier.
  • Hinders upstream movement, requiring more effort.
In a river crossing scenario, being aware of the speed of the current allows for calculating how much it assists or opposes the person's speed. It's often represented as \(v_w\), and is vital for establishing the person's effective speed:
  • Effective downstream speed: \(v_p + v_w\)
  • Effective upstream speed: \(v_p - v_w\)
The speed of the current becomes especially important when determining ratios or solving equations that predict travel outcomes.
River Crossing Problems
River crossing problems involve calculating how a person moves across a river affected by the current. These have real-world applications in navigation, transportation, and even sports.

Such problems often require understanding the effects of current on movement:
  • Identify the speeds involved: both person and current speeds are critical.
  • Calculate effective speeds for upstream and downstream motion accordingly.
  • Use distances traveled or time taken to establish equations that relate these speeds.
In the given problem, the relative speeds affect the path and time it takes to cross the river. Problems like this usually aim to find solutions with the minimum effort from the person involved or to reach a specific target. By determining the minimum speed ratios, like the \(\frac{1}{3}\) answer found in our example problem, you quantify the conditions necessary for successful crossing. Understanding and practicing these components ensures clarity in solving related challenges efficiently.

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Most popular questions from this chapter

Ship \(A\) is travelling with a velocity of \(5 \mathrm{~km} \mathrm{~h}^{-1}\) due east. A second ship is heading \(30^{\circ}\) east of north. What should be the speed of second ship if it is to remain always due north with respect to the first ship? (1) \(10 \mathrm{~km} \mathrm{~h}^{-1}\) (2) \(9 \mathrm{kmh}^{-1}\) (3) \(8 \mathrm{~km} \mathrm{~h}^{-1}\) (4) \(7 \mathrm{~km} \mathrm{~h}^{-1}\)

A ball is thrown at different angles with the same speed \(u\) and from the same point and it has the same range in both the cases. If \(y_{1}\) and \(y_{2}\) are the heights attained in the two cases, then \(y_{1}+y_{2}\) is equal to (1) \(\frac{u^{2}}{g}\) (2) \(\frac{2 u^{2}}{g}\) (3) \(\frac{u^{2}}{2 g}\) (4) \(\frac{u^{2}}{4 g}\)

Two inclined planes \(O A\) and \(O B\) having inclination (with borizontal) \(30^{\circ}\) and \(60^{\circ}\), respectively, intersect each other at \(O\) as shown in figure. A particle is projected from point \(P\) with velocity \(u=10 \sqrt{3} \mathrm{~ms}^{-1}\) along a direction perpendicular to plane \(O A\). If the particle strikes plane \(O B\) perpendicularly at \(Q\), calculate Time of flight of the particle (1) \(8 \mathrm{~s}\) (2) \(6 \mathrm{~s}\) (3) \(4 \mathrm{~s}\) (4) \(2 \mathrm{~s}\)

Aparticle has mitial velocity \(4 i+4 j \mathrm{~m} \mathrm{~s}^{-1}\) and an acceleration \- \(0.4 i \mathrm{~ms}^{-2}\), at what time will its speed be \(5 \mathrm{~ms}^{-1} ?\) (1) \(2.5 \mathrm{~s}\) (2) \(17.5 \mathrm{~s}\) (3) s (4) \(8.5 \mathrm{~s}\)

equation of motion of a projectile is \(y=12 x-\frac{3}{4} x^{2}\). The horizontal component of velocity is \(3 \mathrm{~ms}^{-1}\). What is The hortzo of the projectile? (1) \(18 \mathrm{~m}\) (4) \(21.6 \mathrm{~m}\) (3) \(12 \mathrm{~m}\)
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