/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 119 A swimmer wishes to cross a \(80... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A swimmer wishes to cross a \(800 \mathrm{~m}\) wide river flowing at \(6 \mathrm{~km} / \mathrm{hr}\). His speed with respect to water is \(4 \mathrm{~km} / \mathrm{hr}\). He crosses the river in shortest possible time. He is drifted downstream on reaching the other bank by a distance of (A) \(800 \mathrm{~m}\) (B) \(1200 \mathrm{~m}\) (C) \(400 \sqrt{13} \mathrm{~m}\) (D) \(2000 \mathrm{~m}\)

Short Answer

Expert verified
The swimmer is drifted downstream by 1200 meters on reaching the other bank. Therefore, the correct answer is (B) \(1200 \mathrm{~m}\).

Step by step solution

01

Convert speeds to the same units

Since the width of the river is given in meters, let's convert the swimmer's speed and the river's speed to meters per second (m/s). Speed of river flow: \(\frac{6000 \mathrm m}{3600 \mathrm s} = \frac{5}{3}\mathrm{ms}^{-1}\) Speed of swimmer wrt water: \(\frac{4000 \mathrm m}{3600 \mathrm s} = \frac{10}{9}\mathrm{ms}^{-1}\) Now we have the speed of the river as \(\frac{5}{3}\mathrm{ms}^{-1}\) and the speed of the swimmer wrt water as \(\frac{10}{9}\mathrm{ms}^{-1}\).
02

Find the swimmer's optimal direction

For the swimmer to cross the river in the shortest possible time, he should swim perpendicular to the current. That means the swimmer's speed perpendicular to the river flow is \(\frac{10}{9}\mathrm{ms}^{-1}\) and the same speed can be considered as the resultant speed for the shortest crossing time.
03

Calculate the time taken to cross the river

Now, we will find the time taken to cross the river. We have the swimmer's resultant speed and the river's width, so we can use the formula: \[time = \frac{distance}{speed}\] \(time = \frac{800\mathrm{m}}{\frac{10}{9}\mathrm{ms}^{-1}}= 720\mathrm{s}\) So the swimmer takes 720 seconds to cross the river.
04

Calculate the downstream distance

To find the distance the swimmer is drifted downstream, we can use the river's speed and the time taken to cross the river: Downstream distance = River flow speed * Crossing time Downstream distance = \(\frac{5}{3}\mathrm{ms}^{-1} * 720\mathrm{s} = 1200\mathrm{m}\) The swimmer is drifted downstream by 1200 meters on reaching the other bank. Therefore, the correct answer is: (B) \(1200 \mathrm{~m}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle starts from the origin at \(t=0 \mathrm{~s}\) with a velocity of \(10.0 \hat{j} \mathrm{~m} / \mathrm{s}\) and moves in the \(x-y\) plane with a constant acceleration of \((8.0 \hat{i}+2.0 \hat{j}) \mathrm{ms}^{-2}\) At a time when \(x\)-coordinate of the particle \(16 \mathrm{~m}\), (A) \(24 \mathrm{~m}\) (B) \(22 \mathrm{~m}\) (C) \(20 \mathrm{~m}\) (D) None

The equation of projectile is \(y=\sqrt{3} x-\frac{g x^{2}}{2}\). The angle of projection is (A) \(\theta=\frac{\pi}{6}\) (B) \(\theta=\frac{\pi}{3}\) (C) \(\theta=\frac{\pi}{2}\) (D) \(\theta=\frac{\pi}{12}\)

A car moving along a straight highway with a speed of \(126 \mathrm{~km} \mathrm{~h}^{-1}\) is brought to a stop within a distance of \(200 \mathrm{~m}\). How long does it take for the car to stop? (A) \(5.1 \mathrm{~s}\) (B) \(11.44 \mathrm{~s}\) (C) \(15.2 \mathrm{~s}\) (D) None

A particle is projected with a velocity \(u\), at an angle \(\alpha\), with the horizontal. Time at which its vertical component of velocity becomes half of its net speed at the highest point will be (A) \(\frac{u}{2 g}\) (B) \(\frac{u}{2 g}(\sin \alpha-\cos \alpha)\) (C) \(\frac{u}{2 g}(2 \cos \alpha-\sin \alpha)\) (D) \(\frac{u}{2 g}(2 \sin \alpha-\cos \alpha)\)

A \(2 \mathrm{~m}\) wide truck is moving with a uniform speed \(v_{0}=\) \(8 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed \(v\) when the truck is \(4 \mathrm{~m}\) away from him. The minimum value of \(v\) so that he can cross the road safely is (A) \(2.62 \mathrm{~m} / \mathrm{s}\) (B) \(4.6 \mathrm{~m} / \mathrm{s}\) (C) \(3.57 \mathrm{~m} / \mathrm{s}\) (D) \(1.414 \mathrm{~m} / \mathrm{s}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.