Question

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

Answer 1

Part a)

`A_x = 18.03 km`

`A_y = - 18.03`

`B_x = 21 km`

`B_y = 36.4 km`

Part b)

`R_x = 39.03 km`

`R_y = 18.37 km`

Part c)

`R = 43.13 km`

`\theta = 25.2 degree`North of East

Step-by-step explanation:

Part a)

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

`A_x = 25.5 cos45`

`A_x = 18.03 km`

`A_y = - 25.5 sin45`

`A_y = -18.03 km`

His displacement on second day

`B_x = 42 cos60`

`B_x = 21 km`

`B_y = 42 sin60`

`B_y = 36.4 km`

Part b)

Now hiker's net displacement along East

`R_x = A_x + B_x`

`R_x = 18.03 + 21`

`R_x = 39.03 km`

`R_y = A_y + B_y`

`R_y = -18.03 + 36.4`

`R_y = 18.37 km`

Part c)

Magnitude of net displacement is given as

`R = √(R_x^2 + R_y^2)`

`R = √(18.37^2 + 39.03^2)`

`R = 43.13 km`

direction of the net displacement is given as

`tan\theta = (R_y)/(R_x)`

`tan\theta = (18.37)/(39.03)`

`\theta = 25.2 degree`North of East