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Chapter 3: Problem 17

Use unit vectors to express a displacement of \(120 \mathrm{km}\) at \(29^{\circ}\) counterclockwise from the \(x\) -axis.

Short Answer

Expert verified
The displacement vector expressed in unit vectors is \(104.25 i + 60.89 j \, \mathrm{km}\).

Step by step solution

01

Understand the problem

The problem gives a displacement in polar coordinates, which makes sense in many physics problems. But often, one will want this displacement in Cartesian coordinates. To do this, recall the relationships: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), where \(r\) is the magnitude of displacement and \(\theta\) is the angle it makes with x axis. In terms of the unit vectors \(i\) and \(j\), \(r\cos(\theta)\) is the x component (in the direction of \(i\)) and \(r\sin(\theta)\) is the y component (in the direction of \(j\)).
02

Convert the magnitude from polar to Cartesian coordinates

To convert displacement from polar to Cartesian coordinates, use the trigonometric functions sine and cosine. First, convert the angle from degrees to radians: \(29^{\circ} = 29 \times (\pi/180) = 0.507 \) radians. Calculate the x and y magnitudes of displacement: \(x = 120\,\mathrm{km} \cdot \cos(0.507) = 104.25\,\mathrm{km}\), \(y = 120\,\mathrm{km} \cdot \sin(0.507) = 60.89\,\mathrm{km}\).
03

Write the displacement in terms of unit vectors

Now express the displacement vector in terms of the unit vectors \(i\) and \(j\). This is done by multiplying the magnitudes calculated above by the respective unit vectors: Displacement = \(104.25 i + 60.89 j \, \mathrm{km}\).

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