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Oasis \(A\) is \(90 \mathrm{~km}\) due west of oasis \(B\). A desert camel leaves \(A\) and takes \(50 \mathrm{~h}\) to walk \(75 \mathrm{~km}\) at \(37^{\circ}\) north of due east. Next it takes \(35 \mathrm{~h}\) to walk \(65 \mathrm{~km}\) due south. Then it rests for \(5.0 \mathrm{~h}\). What are the (a) magnitude and (b) direction of the camel's displacement relative to \(A\) at the resting point? From the time the camel leaves \(A\) until the end of the rest period, what are the (c) magnitude and (d) direction of its average velocity and (e) its average speed? The camel's last drink was at \(A ;\) it must be at \(B\) no more than \(120 \mathrm{~h}\) later for its next drink. If it is to reach \(B\) just in time, what must be the (f) magnitude and (g) direction of its average velocity after the rest period?

Short Answer

Expert verified
(a) 62.27 km, (b) 18.26° south of east, (c) 0.692 km/h, (d) 18.26° south of east, (e) 1.56 km/h, (f) 1.004 km/h, (g) due west.

Step by step solution

01

Analyze Initial Path of the Camel

The camel first travels 75 km at an angle of 37° north of due east. We need to decompose this into east and north components using trigonometry. The eastward component is: \[ 75 \times \cos(37^\circ) \approx 59.88 \text{ km} \]The northward component is: \[ 75 \times \sin(37^\circ) \approx 45.14 \text{ km} \]
02

Determine Second Displacement

The second part of the trip involves traveling 65 km due south. This will only affect the north-south component. Since it's due south, the northward component we calculated decreases by 65 km:North component after second leg: \[ 45.14 - 65 = -19.86 \text{ km} \] East component remains the same: 59.88 km.
03

Calculate Camel's Displacement Magnitude

Now we calculate the total displacement from the starting point A:\[ \text{Magnitude} = \sqrt{(59.88)^2 + (-19.86)^2} = \sqrt{3585.14 + 394.36} \approx 62.27 \text{ km} \]
04

Determine Displacement Direction

The direction of the displacement can be calculated using the tangent of the angle:\[ \theta = \tan^{-1}\left(\frac{-19.86}{59.88}\right) \approx -18.26^\circ \]Here, the negative angle means it is south of east.
05

Calculate Total Travel Time Including Rest

The camel travels for 50 hours quickly, followed by 35 hours, then rests for 5 hours, totaling:\[ 50 + 35 + 5 = 90 \text{ hours} \]
06

Find Average Velocity Magnitude and Direction

The average velocity is displacement divided by travel time, so:\[ \text{Average Velocity Magnitude} = \frac{62.27}{90} \approx 0.692 \text{ km/h} \]The direction is the same as the displacement angle, approximately 18.26° south of east.
07

Calculate Average Speed

Average speed is the total path length divided by total travel time. The path length is: Initial leg: 75 km, second leg: 65 km\[ \text{Average Speed} = \frac{75 + 65}{90} = 1.56 \text{ km/h} \]
08

Determine Requirements for Third Segment

The camel has 120 hours in total and has used 90 hours already. The remaining time is:\[ 120 - 90 = 30 \text{ hours} \]Oasis B is 90 km due west of point A, given the camel has traveled 59.88 km east initially, it needs to cover:\[ 90 - 59.88 = 30.12 \text{ km} \] west to reach B.
09

Final Velocity After Rest

To calculate required average velocity after resting:\[ \text{Required Average Velocity} = \frac{30.12}{30} = 1.004 \text{ km/h} \]Since it's due west to reach B, the direction is due west.

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Key Concepts

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

Vector Components
Understanding the idea of breaking down a vector into components is fundamental in physics. When you have a vector, like the camel's initial 75 km journey at 37° north of due east, you're dealing with a direction and a magnitude. In order to simplify motion analysis, we use vector components. These components are like slices of a pie — splitting the vector into more manageable horizontal (east-west) and vertical (north-south) parts.

To find these components, we apply trigonometric functions: the cosine function helps find the horizontal (east) component, and the sine function helps find the vertical (north) component.
  • The eastward component: \[ 75 imes \cos(37^\circ) \approx 59.88 \text{ km} \]
  • The northward component: \[ 75 \times \sin(37^\circ) \approx 45.14 new.\text{ km} \]
These components tell us how far the camel goes in each direction.
Average Speed
The concept of average speed is simply about how much distance the camel covers over a certain amount of time, without concerning itself with direction. It's calculated by taking the total distance traveled and dividing it by the total time taken.

In our exercise, the camel travels a path of 75 km on its first leg and 65 km on its second, totaling 140 km. The total time spent moving, including the rest, is 90 hours:
\[\text{Average Speed} = \frac{75+65}{90} = 1.56 \ \text{km/h}\]
This calculation emphasizes speed, not the actual direction the camel takes.
Average Velocity
Average velocity is slightly different from average speed. While speed covers how much ground is covered, average velocity concerns itself with the change in position from start to end, plus direction. For our camel at rest, it took a total of 90 hours to travel from its start to its resting position.

The displacement, the straight-line measurement from point \(A\) to the point of rest, is the total displacement we calculated using vector components:\[\sqrt{(59.88)^2 + (-19.86)^2} \approx 62.27 \ \text{km}\]
The average velocity is then:\[\text{Average Velocity} = \frac{62.27}{90} \approx 0.692 \ \text{km/h}\]
This also includes direction, approximately 18.26° south of east.
Trigonometry in Physics
Trigonometry serves as a vital tool in physics for analyzing directions and distances. It is extensively used to resolve directions into simpler components. In our scenario with the camel, trigonometry assists in determining the exact path and displacement of the camel's journey.

When the camel moves 75 km at 37° north of east, we used the cosine and sine functions to resolve this movement into horizontal and vertical components, pinpointing how much of this distance is due east and how much shifted north. Further trigonometric calculations like tangent help calculate the angle of resultant displacement:\[\theta = \tan^{-1}\left(\frac{-19.86}{59.88}\right) \approx -18.26^\circ\]
The angle tells us about the direction relative to the east-west line — revealing how far south of east the camel ends up. Such components and angles simplify the analysis of the camel's overall path.

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Most popular questions from this chapter

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