Question

A hiker treks 30 degrees south of east at a speed of 15 m/s for 30 min and then turns due west and hikes at a speed of 8m/s for another 20 min. What is the displacement of this explorer (magnitude and direction)?

Answer 1

Answer:19.3 km,
`\theta =44.40^(\circ)` south of east

Step-by-step explanation:

Given

Hiker treks
`30 ^(\circ)` south of east at a speed of 15 m/s for 30 min and then turns due to west and hikes at speed of 8 m/s for another 20 min

Let position vector of Hiker at the end of 30 min

`r=27000cos30\hat{i}-27000sin30\hat{j}`

after he turns west so new position vector of hiker is

`r'=27000cos30\hat{i}-27000sin30\hat{j}-9600\hat{i}`

`r'=13782.68\hat{i}-13500\hat{j}`

Therefore Displacement is given by |r'|

`|r'|=√(13782.68^2+13500^2)`

`|r'|=19,292.8035 m\ or 19.3 km`

for direction

`tan\theta =(13500)/(13782.68)`

`\theta =44.40^(\circ)` south of east