Question

3. A vector A has a magnitude of 58.0 m and points in a direction 22.0° below the negative x-axis. A second vector, B, has a magnitude of 90.0 m and points in a direction 54.0° below the negative x-axis. Using the component method, find the magnitude of the vector D - A - B x 75.7m 4. Consider the three displacement vectors shown in the figure: Vector A has a magnitude of 8.10 km and a direction that makes an angle θ= 32.0° to the left of the positive y axis, vector B has a magnitude of 6.40 km and a direction that makes an angle of α 30 above the positive x axis. and vector C has a magnitude of 4.40 km and a direction that makes an angle ß = 67.0° below the negative x-axis Determine the magnitude of the vector D-A-B Blank km

Answer 1

3)

$\vec{A}=58m\ at \ 22^{\circ}\ below\ -x\ axis$

Both x and y terms are negative

$\vec{A}=(-A_{x}cos22i-A_{y}sin22j)m$

$\vec{A}=(-58cos22i-58sin22j)m$

${\color{Red} \vec{A}=(-53.78i-21.73j)m}$

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$\vec{B}=90m\ at \ 54^{\circ}\ below\ -x\ axis$

Both x and y terms are negative

$\vec{B}=(-B_{x}cos54i-B_{y}sin54j)m$

$\vec{B}=(-90cos54i-90sin54j)m$

${\color{Red} \vec{B}=(-52.9i-72.81j)m}$

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$\vec{D}=\vec{A}-\vec{B}$

$\vec{D}=(-53.78i-21.73j)-(-52.9i-72.81j)$

$\vec{D}=(-53.78i-21.73j)+(52.9i+72.81j)$

$\vec{D}=(-0.88i+51.08j)m$

$D=\sqrt{(-0.88)^{2}+(51.08)^{2}}$

ANSWER: ${\color{Red} D=51.09m}$

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4).

$\vec{A}=8.1km\ at \ 32^{\circ}\ to\ left\ of\ y\ axis$

x term is negative and y term is positive

$\vec{A}=(-A_{x}sin32i+A_{y}cos32j)m$

$\vec{A}=(-8.1sin32i+8.1cos32j)m$

${\color{Red} \vec{A}=(-4.29i+6.87j)m}$

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$\vec{B}=6.4km\ at \ 30^{\circ}\ above\ x\ axis$

Both x and y terms are positive

$\vec{B}=(B_{x}cos30i+B_{y}sin30j)m$

$\vec{B}=(6.4cos30i+6.4sin30j)m$

${\color{Red} \vec{B}=(5.54i+3.2j)m}$

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$\vec{C}=4.4km\ at \ 67^{\circ}\ below\ -x\ axis$

Both x and y terms are negative

$\vec{C}=(-C_{x}cos67i-C_{y}sin67j)m$

$\vec{C}=(-4.4cos67i-4.4sin67j)m$

${\color{Red} \vec{C}=(-1.72i-4.05j)m}$

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$\vec{D}=\vec{A}-\vec{B}+\vec{C}$

$\vec{D}=(-4.29i+6.87j)-(5.54i+3.2j)+(-1.72i-4.05j)$

$\vec{D}=(-4.29i+6.87j)+(-5.54i-3.2j)+(-1.72i-4.05j)$

$\vec{D}=(-4.29-5.54-1.72)i+(6.87-3.2-4.05)j$

$\vec{D}=-11.55i-0.38j$

$\vec{D}=\sqrt{(-11.55)^{2}+(-0.38j)^{}}$

ANSWER: ${\color{Red} \vec{D}=11.56km}$

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