You are to make four straight-line moves over a flat desert floor,starting at th

Question

You are to make four straight-line moves over a flat desert floor,starting at the origin of an xy coordinate system andending at the xy coordinates (-160 m, 36 m). The xcomponent and y component of your moves are the following,respectively, in meters: (20 and 60), then (bx and -70),then (-20 and cy), then (-65 and-74). (a) What is component bx?
1 m

(b) What is component cy?
2 m

(c) What is the the magnitude of the overall displacement?
3 m

(d) What is the angle of the overall displacement?
4° (counterclockwise from the positivex axis) (a) What is component bx?
1 m

(b) What is component cy?
2 m

(c) What is the the magnitude of the overall displacement?
3 m

(d) What is the angle of the overall displacement?
4° (counterclockwise from the positivex axis)

Explanation / Answer

In vector notation : ------------------ Initial position r = 0 i + 0 j final position r ' = -160 i + 36 j Overall displacement S = r ' - r                                     = -160 i + 36 j is t displacement   s = 20 i + 60 j 2 nd displacement s ' = bx i -70j 3 rd displacement s " = -20 i + cyj 4 th displacement s "' = -65i -74 j we know S = s + s' +s" + s"' -160 i + 36 j = ( 20+bx-20-65 ) i + ( 60-70+cy-74 ) j                    = (bx-65) i + ( -84 + cy ) j from this bx -65 = -160                    bx = -95 m -84 + cy = 36             c y = 120 m (c).  the magnitude of the overall displacement =modulus of S                                                                      = [ -160^2 +36 ^ 2 ]                                                                      = 164 m (d). Let the overall displacement S makes an angle with negative x - axis then tan = 36 /160 from this = 12.68 degrees So, angle with positive x- axis = 180 + = 192.68degrees we know S = s + s' +s" + s"' -160 i + 36 j = ( 20+bx-20-65 ) i + ( 60-70+cy-74 ) j                    = (bx-65) i + ( -84 + cy ) j from this bx -65 = -160                    bx = -95 m -84 + cy = 36             c y = 120 m (c).  the magnitude of the overall displacement =modulus of S                                                                      = [ -160^2 +36 ^ 2 ]                                                                      = 164 m (d). Let the overall displacement S makes an angle with negative x - axis then tan = 36 /160 from this = 12.68 degrees So, angle with positive x- axis = 180 + = 192.68degrees

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