Ax+Ay≠A (3.5)
If the vectorAis known, then its magnitudeA(its length) and its angleθ(its direction) are known. To findAxandAy, itsx- andy-components,
we use the following relationships for a right triangle.Ax=Acosθ (3.6)
andAy=Asinθ. (3.7)
Figure 3.27The magnitudes of the vector componentsAxandAycan be related to the resultant vectorAand the angleθwith trigonometric identities. Here we see
thatAx=AcosθandAy=Asinθ.
Suppose, for example, thatAis the vector representing the total displacement of the person walking in a city considered inKinematics in Two
Dimensions: An IntroductionandVector Addition and Subtraction: Graphical Methods.Figure 3.28We can use the relationshipsAx=AcosθandAy=Asinθto determine the magnitude of the horizontal and vertical component vectors in this
example.ThenA= 10.3blocks andθ= 29.1º, so that
Ax=Acosθ=⎛⎝10.3 blocks⎞⎠⎛⎝cos 29.1º⎞⎠= 9.0 blocks (3.8)
A (3.9)
y=Asinθ=
⎛⎝10.3 blocks
⎞
⎠⎛⎝sin 29.1º
⎞⎠= 5.0 blocks.
Calculating a Resultant Vector
If the perpendicular componentsAxandAyof a vectorAare known, thenAcan also be found analytically. To find the magnitude Aand
directionθof a vector from its perpendicular componentsAxandAy, we use the following relationships:
A= A (3.10)
x^2 +Ay^2
θ= tan−1(A (3.11)
y/Ax).
96 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS
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