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Use the worked example above to help you solve this problem. A hiker begins a trip by first walking 25.5 km southeast from her base camp. On the second day she walks 42.0 km in a direction 60.0° north of east, at which point she discovers a forest ranger's tower. (a) Determine the components of the hiker's displacements in the first and second days. Ax = km Ay = km Bx = km By = km (b) Determine the components of the hiker's total displacement for the trip. Rx = km Ry = km (c) Find the magnitude and direction of the displacement from base camp. Magnitude km Direction ° north of east

Respuesta :

Answer:

Part a)

[tex]A_x = 18.03 km[/tex]

[tex]A_y = - 18.03[/tex]

[tex]B_x = 21 km[/tex]

[tex]B_y = 36.4 km[/tex]

Part b)

[tex]R_x = 39.03 km[/tex]

[tex]R_y = 18.37 km[/tex]

Part c)

[tex]R = 43.13 km[/tex]

[tex]\theta = 25.2 degree[/tex]North of East

Explanation:

Part a)

Hiker's displacement on first day of the motion is given as

[tex]A_x = 25.5 cos45[/tex]

[tex]A_x = 18.03 km[/tex]

[tex]A_y = - 25.5 sin45[/tex]

[tex]A_y = -18.03 km[/tex]

His displacement on second day

[tex]B_x = 42 cos60[/tex]

[tex]B_x = 21 km[/tex]

[tex]B_y = 42 sin60[/tex]

[tex]B_y = 36.4 km[/tex]

Part b)

Now hiker's net displacement along East

[tex]R_x = A_x + B_x[/tex]

[tex]R_x = 18.03 + 21[/tex]

[tex]R_x = 39.03 km[/tex]

[tex]R_y = A_y + B_y[/tex]

[tex]R_y = -18.03 + 36.4[/tex]

[tex]R_y = 18.37 km[/tex]

Part c)

Magnitude of net displacement is given as

[tex]R = \sqrt{R_x^2 + R_y^2}[/tex]

[tex]R = \sqrt{18.37^2 + 39.03^2}[/tex]

[tex]R = 43.13 km[/tex]

direction of the net displacement is given as

[tex]tan\theta = \frac{R_y}{R_x}[/tex]

[tex]tan\theta = \frac{18.37}{39.03}[/tex]

[tex]\theta = 25.2 degree[/tex]North of East

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