A pilot in a small plane encounters shifting winds. He flies 30.0 km at 30 north

Question

A pilot in a small plane encounters shifting winds. He flies 30.0 km at 30 north of east, then 50.0 km due north. From this point, he flies an additional distance in an unknown direction, only to find himself at a small airstrip that his map shows to be 90.0 km directly north of his starting point.

Part A

What was the length of the third leg of his trip?

Part B

What was the direction of the third leg of his trip? Do not use the standard angle. Instead give the direction as how many degrees north of west.

Explanation / Answer

Here ,

let east is positive x - direction

north is positive y-direction

let the unknown displacement is d3

a)

as d1 + d2 + d3 = dnet

30 *(cos(30) i + j *sin(30)) + 50 j + d3 = 90 j

solving for d3

d3 = - 26 i + 25 j km

length of d3 = sqrt(26^2 + 25^2)

length of d3 = 36.1 km

the length of third leg of trip is 36.1 km

b)

direction = arctan(25/26) north of west

direction = 43.9 degree north of west

the direction of the third leg of his trip is 43.9 degree north of west

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