Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 6

A ship sets out to sail to a point \(124 \mathrm{~km}\) due north. An unexpected storm blows the ship to a point \(72.6 \mathrm{~km}\) to the north and \(31.4 \mathrm{~km}\) to the east of its starting point. How far, and in what direction, must it now sail to reach its original destination?

Short Answer

Expert verified
The ship must sail a distance equal to the magnitude of the resultant vector C, and in the direction of this vector from the north (measured in a clockwise direction).

Step by step solution

01

Identify Given Vectors

From the problem, identify the given vectors. The initial intended path represents Vector A that is 124 km due north. The storm has caused a deviation which can be represented by Vector B, with two components - 72.6 km to the north and 31.4 km to the east.
02

Determine Resultant Vector from Storm

Vector B, the resultant vector from the storm, will contain both components - the northward and eastward distances travelled. Vector B will have a direction (angle \(\theta_B\)) to north. Using trigonometry, we get \(\theta_B = tan^{-1}( E_{B} / N_{B})\) where \(E_{B}\) is the eastward distance (31.4 km) and \(N_{B}\) is the northward distance (72.6 km). The magnitude of vector B can be calculated by Pythagorean theorem as \( \|B\| = \sqrt{(N_{B}^2 + E_{B}^2)} \)
03

Determine Required Vector C

The ship must now sail along a path that could be represented by Vector C to reach its original destination. We know that C = A - B. Since A is purely in north direction, \(N_{A}\) = A = 124 km, \(E_{A}\) = 0. So, we can find the northward and eastward components of Vector C as subtracting respective components of Vector B from A. Hence, \( N_{C} = N_{A} - N_{B}\), \(E_{C} = E_{A} - E_{B}\). Let's compute these.
04

Calculate Magnitude and Direction of C

The magnitude or how far the ship needs to sail (\( \|C\| \)) can be found by applying the Pythagorean theorem to the northward (\(N_{C}\)) and eastward (\(E_{C}\)) components of Vector C. Hence, \( \|C\| = \sqrt{(N_{C}^2 + E_{C}^2)} \). The direction of Vector C or the direction in which the ship must sail can be determined by taking the arctangent of its eastward distance over its northward distance. Hence, \(\theta_{C} = tan^{-1} (E_{C} / N_{C}) \)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The position of a particle moving in an \(x y\) plane is given by \(\overrightarrow{\mathbf{r}}=\left[\left(2 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}-(5 \mathrm{~m} / \mathrm{s}) t\right] \hat{\mathbf{i}}+\left[(6 \mathrm{~m})-\left(7 \mathrm{~m} / \mathrm{s}^{4}\right) t^{4}\right] \hat{\mathbf{j}} . \quad\) Cal- culate \((a) \overrightarrow{\mathbf{r}},(b) \overrightarrow{\mathbf{v}}\), and \((c) \overrightarrow{\mathbf{a}}\) when \(t=2 \mathrm{~s}\).

An automobile traveling \(35 \mathrm{mi} / \mathrm{h}(=56 \mathrm{~km} / \mathrm{h})\) is \(110 \mathrm{ft}(=34\) \(\mathrm{m}\) ) from a barrier when the driver slams on the brakes. Four seconds later the car hits the barrier. ( \(a\) ) What was the automobile's constant deceleration before impact? (b) How fast was the car traveling at impact?

A jumbo jet needs to reach a speed of \(360 \mathrm{~km} / \mathrm{h}(=224 \mathrm{mi} / \mathrm{h})\) on the runway for takeoff. Assuming a constant acceleration and a runway \(1.8 \mathrm{~km}(=1.1 \mathrm{mi})\) long, what minimum acceleration from rest is required?

The velocity of a particle moving in the \(x y\) plane is given by \(\overrightarrow{\mathbf{v}}=\left[\left(6.0 \mathrm{~m} / \mathrm{s}^{2}\right) t-\left(4.0 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\right] \hat{\mathbf{i}}+(8.0 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\). Assume \(t>0 .(a)\) What is the acceleration when \(t=3 \mathrm{~s} ?(b)\) When (if ever) is the acceleration zero? \((c)\) When (if ever) is the velocity zero? \((d)\) When (if ever) does the speed equal \(10 \mathrm{~m} / \mathrm{s}\) ?

A rock is dropped from a 100 -m-high cliff. How long does it take to fall \((a)\) the first \(50.0 \mathrm{~m}\) and \((b)\) the second \(50.0 \mathrm{~m}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Mechanics and Materials

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Turning Points in Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Waves Physics

Read Explanation
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.

Sign-up for free