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(17%) Problem 5: I start walking. The 1st leg of my trip I walk dA = 65 m at ,-7

ID: 1777740 • Letter: #

Question

(17%) Problem 5: I start walking. The 1st leg of my trip I walk dA = 65 m at ,-78 south of east. The 2nd leg of my trip I walk dg-115 m at 0g = 17° north of east. On my final leg I walk dc = 75 m at 0c = 66.5° north of west. Choose the coordinate system so that x is directed towards the east, and y is directed towards the north G 25% Part (a) Write an expression for the x-component of the final displacement in terms of the given quantities Grade Summar Potential cos() cos(%) sin() sin(%) cos() cos(%) sin() sin(%) cos(0) cos(9C) sin() sin(e) | | Submissions Attempts remain (5% per attempt detailed view 123 | | 0 END BACKSPACE DELCLEAR Submit Hint I give up! Hints: 3% deduction per hint. Hints remaining: 3 Feedback: 3% deduction per feedback. D 25% Part (b) Write an expression for the y-component of the final displacement in terms of the given quantities D 25% Part (c) What is the magnitude of my displacement vector (in meters) as measured from the origin? D 25% Part (d) What is the angle of my displacement vector as measured counterclockwise from the +x-axis?

Explanation / Answer

given, 1st leg of the trip

da = 65 m, thetaa = 7 deg south of east

assuming i and j be unit vectors along east and north respectively

in vector notation

da = 65(cos(7)i - sin(7)j)

leg2

db = 115 m, thetab = 17 north of east

db = 115(cos(17)i + sin(17)j)

leg 3

dc = 75 m, thetac = 66.5 north of west

dc = 75(-cos(66.5)i + sin(66.5)j)

a. final x displacement = dacos(theta) + dbcos(thetab) - dccos(thetac)

b. final y displacement = -dasin(theta) + dbsin(tehtab) + dcsin(thetac)

c. final x displacement = 65*cos(7) + 115cos(17) - 75cos(66.5) = 144.584 m

final y displacement = -65sin(7) + 115sin(17) + 75sin(66.5) = 94.48 m

final displacement from origin = sqroot(x^2 + y^2) = 172.7169 m

d. angle of this displacement counterclockwise form +x axis = theta

tan(theta) = y/x

theta = 33.163 deg