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Chapter 3: Problem 1

Displacement vector \(\overrightarrow{\mathbf{A}}\) is directed to the west and has magnitude \(2.56 \mathrm{km} .\) A second displacement vector is also directed to the west and has magnitude \(7.44 \mathrm{km}\) (a) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} ?\) (b) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}} ?\) (c) What are the magnitude and direction of \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}} ?\)

Short Answer

Expert verified
Question: Calculate the magnitudes and directions of the following vector operations, given \(\overrightarrow{\mathbf{A}} = -2.56 \mathrm{km}\) west and \(\overrightarrow{\mathbf{B}} = -7.44 \mathrm{km}\) west: (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) (c) \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}\) Answer: (a) Magnitude: \(10 \mathrm{km}\), Direction: West (b) Magnitude: \(4.88 \mathrm{km}\), Direction: East (c) Magnitude: \(4.88 \mathrm{km}\), Direction: West

Step by step solution

01

(a) Calculate \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\)

Since both vectors are directed to the west, we can represent them as negative scalar values: \(\overrightarrow{\mathbf{A}} = -2.56 \mathrm{km}\) and \(\overrightarrow{\mathbf{B}} = -7.44 \mathrm{km}\) Adding them together, \(\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} = (-2.56) + (-7.44) = -10 \mathrm{km}\) The magnitude is \(10 \mathrm{km}\), and as the result is negative, the direction is to the west.
02

(b) Calculate \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\)

Subtracting the vectors, \(\overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} = (-2.56) - (-7.44) = 4.88 \mathrm{km}\) The magnitude is \(4.88 \mathrm{km}\), and as the result is positive, the direction is to the east.
03

(c) Calculate \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}\)

Subtracting the vectors in reverse order, \(\overrightarrow{\mathbf{B}} - \overrightarrow{\mathbf{A}} = (-7.44) - (-2.56) = -4.88 \mathrm{km}\) The magnitude is \(4.88 \mathrm{km}\), and as the result is negative, the direction is to the west.

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Most popular questions from this chapter

The range \(R\) of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at \(t=0\) with initial speed \(v_{i}\) at an angle \(\theta\) above the horizontal. (a) Find the time \(t\) at which the projectile returns to its original altitude. (b) Show that the range is \(R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}\) [Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$ (c) What value of \(\theta\) gives the maximum range? What is this maximum range?
A speedboat heads west at \(108 \mathrm{km} / \mathrm{h}\) for 20.0 min. It then travels at \(60.0^{\circ}\) south of west at \(90.0 \mathrm{km} / \mathrm{h}\) for 10.0 min. (a) What is the average speed for the trip? (b) What is the average velocity for the trip?
The citizens of Paris were terrified during World War I when they were suddenly bombarded with shells fired from a long-range gun known as Big Bertha. The barrel of the gun was \(36.6 \mathrm{m}\) long and it had a muzzle speed of \(1.46 \mathrm{km} / \mathrm{s} .\) When the gun's angle of elevation was set to \(55^{\circ},\) what would be the range? For the purposes of solving this problem, neglect air resistance. (The actual range at this elevation was \(121 \mathrm{km} ;\) air resistance cannot be ignored for the high muzzle speed of the shells.)
You want to make a plot of the trajectory of a projectile. That is, you want to make a plot of the height \(y\) of the projectile as a function of horizontal distance \(x\). The projectile is launched from the origin with an initial speed \(v_{\mathrm{i}}\) at an angle \(\theta\) above the horizontal. Show that the equation of the trajectory followed by the projectile is $$y=\left(\frac{v_{\text {iy }}}{v_{\text {ix }}}\right) x+\left(\frac{-g}{2 v_{\text {ix }}^{2}}\right) x^{2}$$
At \(t=0,\) an automobile traveling north begins to make a turn. It follows one- quarter of the arc of a circle with a radius of \(10.0 \mathrm{m}\) until, at \(t=1.60 \mathrm{s},\) it is traveling east. The car does not alter its speed during the turn. Find (a) the car's speed, (b) the change in its velocity during the turn, and (c) its average acceleration during the turn.
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