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# Problem: I start walking. The 1st leg of my trip I walk dA = 105 m at θA = 24° south of east. The 2nd leg of my trip I walk dB = 65 m at θB = 28° north of east. On my final leg I walk dC = 95 m at θC = 63° north of west. Take east to be the +x direction and north to be the +y direction.Part (a) Write an expression for the x component of the final displacement in terms of the given quantities.Part (b) Write an expression for the y component of the final displacement in terms of the given quantities.Part (c) What is the magnitude of my displacement vector (in meters) as measured from the origin?Part (d) What is the angle of my displacement vector as measured counterclockwise from the +x-axis (East)?

###### FREE Expert Solution

Let east be the positive x-axis and north be the positive y-axis.

Vector components:

Vector magnitude:

$\overline{)\mathbf{|}\stackrel{\mathbf{⇀}}{\mathbit{A}}\mathbf{|}{\mathbf{=}}\sqrt{{{\mathbit{A}}_{\mathbit{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbit{A}}_{\mathbit{y}}}^{\mathbf{2}}}}$

Direction:

$\overline{){\mathbf{tan}}{\mathbit{\theta }}{\mathbf{=}}\frac{{\mathbit{A}}_{\mathbit{y}}}{{\mathbit{A}}_{\mathbit{x}}}}$

Part (a)

All angles should be measured from the positive x-axis in a counterclockwise direction.

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###### Problem Details

I start walking. The 1st leg of my trip I walk dA = 105 m at θA = 24° south of east. The 2nd leg of my trip I walk dB = 65 m at θB = 28° north of east. On my final leg I walk dC = 95 m at θC = 63° north of west. Take east to be the +x direction and north to be the +y direction.

Part (a) Write an expression for the x component of the final displacement in terms of the given quantities.

Part (b) Write an expression for the y component of the final displacement in terms of the given quantities.

Part (c) What is the magnitude of my displacement vector (in meters) as measured from the origin?

Part (d) What is the angle of my displacement vector as measured counterclockwise from the +x-axis (East)?