Question

In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day, and she is blown along the following straight lines: 2.50 km and 45.0° north of west, then 4.70 km and 60.0° south of east, then 1.30 km and 25.0° south of west, then 5.10 km due east, then 1.70 km and 5.00° east of north, then 7.20 km and 55.0° south of west, and finally 2.80 km and 10.0° north of east. Use the analytical method to find the resultant vector of all her displacement vectors. What is its magnitude and direction?

Answer 1

Answer:

R = 7,34 km

Ф = 63,43° north of east

Step-by-step explanation:

Let´s analyze each movement and find its perpendicular components. Then, we are going to add the components along the x and y axes to get the resultant and finally to calculate its magnitude and direction.

To determine the perpendicular components, we will use right triangles trigonometrics ratios.

Please see images for each displacement in attached file.

• Displacement A: 2.50 km and 45° north of west

Ax= AcosФ = 2.50cos45° = 1.77 km (neg. direction)

Ay= AsenФ = 2.50sen45° = 1.77 km (pos. direction)

• Displacement B: 4.70 km and 60° south of east

Bx= BcosФ = 4.70cos60° = 2,35 km (pos. direction)

By= BsenФ = 4.70sen60° = 4,07 km (neg. direction)

• Displacement C: 1.30 km and 25° south of west

Cx= CcosФ = 1.30cos25° = 1,18 km (neg. direction)

Cy= CsenФ = 1.30sen25° = 0.55 km (neg. direction)

• Displacement D: 5.10 km due east (0°)

Dx= DcosФ = 5.10 km (pos. direction)

Dy= DsenФ = 0 km

• Displacement E: 1.70 km and 5° east of north

Ex= EsenФ = 1.70sen5° = 0,15 km (pos. direction)

Ey= EcosФ = 1.70cos5° = 1.69 km (pos. direction)

• Displacement F: 7.20 km and 55° south of west

Fx= FcosФ = 7.20cos55° = 4.13 km (neg. direction)

Fy= FsenФ = 7.20sen55° = 5.90 km (neg. direction)

• Displacement G: 2.80 km and 10° north of east

Gx= GcosФ = 2.80cos10° = 2.76 km (pos. direction)

Gy= GsenФ = 2.80sen10° = 0.49 km (pos. direction)

Now, we add the components along the x- and y- axis to find the components of the resultant (R):

Rx = Ax + Bx + Cx + Dx + Ex + Fx + Gx

Ry = Ay + By + Cy + Dy + Ey + Fy + Gy

Therefore,

Rx= (-1.77)+(2.35)+(-1.18)+(5.10)+(0.15)+(-4.13)+(2.76)

Ry= (1.77)+(-4.07)+(-0.55)+(0)+(1.69)+(-5.90)+(0.49)

Note: minus (-) symbol to negative directions

Rx = 3.28 km

Ry = - 6.57 km

Let´s use the Theorem of Pythagoras to find the magnitud ot the resultant:

$R = \sqrt{Rx^{2} +Ry^{2} }$

$R =\sqrt{(3.28)^{2}+(-6.57)^{2} }$

R = 7,34 km

To find the direction of the resultant:

tanФ = $\frac{Ry}{Rx}$

tanФ = $\frac{6.57}{3.28}$

tanФ = 2.00

Ф = tan-1 (2.00)

Ф = 63,43° north of east

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