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Chapter 3: Problem 8

A jogger runs \(100 \mathrm{~m}\) due west, then changes direction for the second leg of the run. At the end of the run, she is 175 \(\mathrm{m}\) away from the starting point at an angle of \(15.0^{\circ}\) north of west. What were the direction and length of her second displacement? Use graphical techniques.

Short Answer

Expert verified
The length of the jogger's second displacement is the value obtained at step 4, and the direction north of west is the value obtained at step 5.

Step by step solution

01

Draw the Initial Vector

Draw a vector \(\vec{A}\) 100m long representing the first leg of the jog. This vector points due west.
02

Draw the Resultant Vector

Draw the resulting vector \(\vec{R}\) 175m long coming from the origin of \(\vec{A}\) and pointing 15掳 north of west. The angle is measured in anticlockwise direction from the west (left direction) towards the direction of the resultant vector if we start from the end of vector \(\vec{A}\).
03

Draw the Second Vector

The second displacement \(\vec{B}\) can be represented as the vector which starts from the end of vector \(\vec{A}\) and leads to the end of \(\vec{R}\). Thus, it forms a triangle with vectors \(\vec{A}\) and \(\vec{R}\). This vector \(\vec{B}\) represents the second leg of the jog.
04

Calculate the Magnitude of the Second Displacement

The magnitude of the second displacement can be calculated using the Pythagorean theorem. Let \(B\) be the magnitude of the second displacement. So, we have \(B^2 = R^2 - A^2\), where \(R = 175m\) is the resultant displacement and \(A = 100m\) is the first displacement. Solving this gives us the magnitude of \(B\).
05

Calculate the Direction of the Second Displacement

The direction \(\theta\) of the second displacement (measured anticlockwise from due west) can be calculated using trigonometry. In triangle OAB (formed by vectors \(\vec{O}\), \(\vec{A}\), and \(\vec{B}\)), we know two sides and one angle. So, we can use the law of cosines to find angle \(\theta\), which is \(cos^{-1}((A^2 + B^2 - R^2) / (2AB))\). Solving this gives us the direction of the second displacement.

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Key Concepts

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

Vector Addition
Vector addition is a fundamental concept in physics that deals with combining two or more vectors into a single resultant vector. When a jogger runs in one direction and then changes course, like in the given exercise, the total displacement can be determined by vector addition. To visualize this, draw arrows representing each leg of the jog, with the length of the arrow proportional to the distance covered and the direction pointing the way the jogger ran. The arrow representing the second leg of the run starts where the first one ends. The combined effect of these two vectors is the jogger's overall displacement from the starting point.
Graphical Vector Analysis
Graphical vector analysis allows us to visually determine the magnitude and direction of a resultant vector. In the exercise, the jogger's path is represented by two vectors laid end-to-end, forming a triangle with the start and finish points. By drawing a straight line from the starting point to the final position, we identify the resultant vector. The length and orientation of this vector provide the total displacement. For greater precision, a scale can be used, where one unit length on the paper might represent a certain number of meters in the jogger's run. This technique is an intuitive way to understand and solve vector problems.
Pythagorean Theorem
The Pythagorean theorem is a principle in geometry that helps to solve for lengths in right-angled triangles. It states that in such a triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In our jogger's displacement problem, the path taken can form a right-angled triangle where the first leg of the run and the resultant displacement are two sides of the triangle, with the second leg being the hypotenuse. By applying this theorem, we can calculate the length of the second leg, which is the second part of the jogger's displacement.
Trigonometry in Physics
Trigonometry plays a crucial role in physics, particularly in problems involving angles and lengths like the jogger's displacement. It involves the study of relationships between the angles and sides of triangles. Trigonometric functions like sine, cosine, and tangent can be used to solve for unknown sides or angles. In our exercise, trigonometry is applied through the law of cosines to calculate the direction of the jogger's second displacement as it is not a right-angled triangle.
Law of Cosines
The law of cosines is a useful rule in trigonometry for determining a particular side or angle within any type of triangle, extending beyond right-angled triangles. It states that the square of one side is equal to the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the included angle. In the context of the jogger's problem, once we know two sides and the included angle thanks to the initial displacement and resultant vector, we can apply the law of cosines to find the second angle and, by extension, the precise direction of the jogger's second leg.

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Most popular questions from this chapter

A jet airliner moving initially at \(3.00 \times 10^{2} \mathrm{mi} / \mathrm{h}\) due east enters a region where the wind is blowing \(1.00\) \(\times 10^{2} \mathrm{mi} / \mathrm{h}\) in a direction \(30.0^{\circ}\) north of east. (a) Find the components of the velocity of the jet airliner relative to the air, \(\vec{v}_{\mu}\). (b) Find the components of the velocity of the air relative to Earth, \(\overrightarrow{\mathbf{v}}_{M \mathcal{E}}\). (c) Write an equation analo- gous to Equation \(3.16\) for the velocities \(\vec{v}_{j}, \vec{v}_{A B}\), and \(\vec{v}_{g E}\) (d) What is the speed and direction of the aircraft relative to the ground?

A man lost in a maze makes three consecutive displacements so that at the end of his travel he is right back where he started. The first displacement is \(8.00 \mathrm{~m}\) westward, and the second is \(13.0 \mathrm{~m}\) northward. Use the graphical method to find the magnitude and direction of the third displacement.

A figure skater glides along a circular path of radius \(5.00 \mathrm{~m}\). If she coasts around one half of the circle, find (a) the magnitude of the displacement vector and (b) what distance she skated. (c) What is the magnitude of the displacement if she skates all the way around the circle?

A bomber is flying horizontally over level terrain at a speed of \(275 \mathrm{~m} / \mathrm{s}\) relative to the ground and at an altitude of \(3.00 \mathrm{~km}\). (a) The bombardier releases one bomb. How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance. (b) Firing from the people on the ground suddenly incapacitates the bombardier before he can call. "Bombs away!" Consequently, the pilot maintains the plane's original course, altitude, and speed through a storm of flak, Where is the plane relative to the bomb's point of impact when the bomb hits the ground? (c) The plane has a telescopic bombsight set so that the bomb hits the target seen in the sight at the moment of release. At what angle from the vertical was the bombsight set?

The record distance in the sport of throwing cowpats is \(81.1 \mathrm{~m}\). This record toss was set by Steve Urner of the United States in 1981 . Assuming the initial launch angle was \(45^{\circ}\) and neglecting air resistance, determine (a) the initial speed of the projectile and (b) the total time the projectile was in flight. (c) Qualitatively, how would the answers change if the launch angle were greater than \(45^{\circ}\) ? Explain.
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