Example 3.2 A Vacation Trip (a) Graphical method for finding the resultant displ

Question

Example 3.2 A Vacation Trip (a) Graphical method for finding the resultant displacement vector = +. (b) Adding the vectors in reverse order ( + ) gives the same result for . A car travels 14.0 km due north and then 34.0 km in a direction 60.0? west of north as shown in figure (a). Find the magnitude and direction of the car's resultant displacement. SOLVE IT Conceptualize The vectors and drawn in figure (a) help us conceptualize the problem. Categorize We can categorize this example as a simple analysis problem in vector addition. The displacement is the resultant when the two individual displacements and are added. We can further categorize it as a problem about the analysis of triangles, so we appeal to our expertise in geometry and trigonometry. Analyze In this example, we show two ways to analyze the problem of finding the resultant of two vectors. The first way is to solve the problem geometrically, using graph paper and a protractor to measure the magnitude of and its direction in figure (a). (In fact, even when you know you are going to be carrying out a calculation, you should sketch the vectors to check your results.) With an ordinary ruler and protractor, a large diagram typically gives answers to two-digit but not to three-digit precision. The second way to solve the problem is to analyze it algebraically. The magnitude of can be obtained from the law of cosines as applied to the triangle. Use R2 = A2 + B2 ? 2AB cos ? from the law of cosines to find R: R = A2 + B2 ? 2AB cos(?) Substitute numerical values, noting that ? = 180? ? 60? = 120?: R = (14.0)2 + (34.0)2 ? 2(14.0)(34.0)cos(120?) R = km Use the law of sines to find the direction of measured from the northerly direction. sin ? B = sin ? R sin ? = B R sin ? = 29.4 km R ? = ? The resultant displacement of the car is at a distance R in a direction ? west of north. Finalize Does the angle ? that we calculated agree with an estimate made by looking at figure (a) or with an actual angle measured from the diagram using the graphical method? Is it reasonable that the magnitude of is larger than that of both and ? Are the units of correct? Although the graphical method of adding vectors works well, it suffers from two disadvantages. First, some people find using the laws of cosines and sines to be awkward. Second, a triangle only results if you are adding two vectors. If you are adding three or more vectors, the resulting geometric shape is usually not a triangle. MASTER IT HINTS: GETTING STARTED | I'M STUCK! A crow flies in two successive displacements to a point that is 80 m to the west. Its first displacement is 80 m in a direction ?1 = 40? west of north. (a) What is the magnitude of its second displacement? m (b) What is its direction of its second displacement measured by the angle ?2 measured west from north? ?

Explanation / Answer

Horizontal component of R is = B Sin 60 = 34 Sin 60 = 29.445 km

Vertical component of R is = A + B Cos 60 = 14 + 34 Cos 60 = 31 km

Resultant R = (29.4452 + 312) = 42.755 km

Its direction from horizontal = atan (Vertical component of R / Horizontal component of R) = atan(31/29.445) = 46.47 deg

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For crow problem:

Let the magnitude of second displacement be B.

We have for horizontal component, 80 = 80 Sin 40 + B Sin (180 - 2)..................(1)

and for vertical component, 0 = 80 Cos 40 - B Cos (180 - 2)..........................(2)

From (1), B Sin (180 - 2) = 28.577.................(3)

From (2), B Cos (180 - 2) = 61.284......................(4)

Squaring both these equations and adding, B2 [ Sin2(180-2) + Cos2(180-2) ] = 4572.319

or, B2 = 4572.319

(a) So, B = 67.619 m

(b) Dividing (4) from (3), Tan (180 - 2) = 28.577/61.284 = 0.466

Thus, (180 - 2) = 25 deg

Thus, 2 = 155 deg

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