Asked by Mary
A hiker walks 10.0 km [n], 6.8km [w], 2.4km [s] and finally 1.2km [w] in 2.5hrs.
a) What is the hiker's displacement
b) In what direction must the hiker set out, in order to return by the most direct route to the starting point?
c)if the hiker walks at a constant speed for the entire trip and returns by the most direct route, how long will the total walk take
a) What is the hiker's displacement
b) In what direction must the hiker set out, in order to return by the most direct route to the starting point?
c)if the hiker walks at a constant speed for the entire trip and returns by the most direct route, how long will the total walk take
Answers
Answered by
Anonymous
north 10 - 2.4 = 7.6
west 6.8 + 1.2 = 8
length of distance vector= displacement = sqrt(64+57.8) = 11.0
direction from start:
tan (North/West) = 7.6 /8 = 0.95
so angle north of west = 43.5 deg which is 270 + 43.5 = 313.5 degrees clockwise from north on compass
so to head back home go 11 miles at 313.5-180 = 133.5 compass direction
total walk = 7.6 + 8 + 11
west 6.8 + 1.2 = 8
length of distance vector= displacement = sqrt(64+57.8) = 11.0
direction from start:
tan (North/West) = 7.6 /8 = 0.95
so angle north of west = 43.5 deg which is 270 + 43.5 = 313.5 degrees clockwise from north on compass
so to head back home go 11 miles at 313.5-180 = 133.5 compass direction
total walk = 7.6 + 8 + 11
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