A ship sails from Guam and travels d1=290 km at theta1=40.0 degrees North of Wes

Question

A ship sails from Guam and travels d1=290 km at theta1=40.0 degrees North of West. But then, the navigation equipment fails. The captain realizes that a small adjustment of d3=15.0 km at theta3=30 West of South is needed to once again reach the goal of being 110 km due East of Guam. Determine the magnitude and direction of that second displacement, d2 and theta2.

I can solve the problem if the 'equipment failure' wouldn't occur. If I exclude the 'equipment failure' I get an answer of d=217.6 km and theta62.5 degrees South of East.

The way I solved the problem without the chance of 15km and 30 degrees W of S is like this

KNOWN= 290 km at 40 degrees, and that you want to be 110 km East of Guam.

cos40=x/290

=x=290cos40

=222.2i

sin40=y/290

=y=290sin40

=186.4j

110i-222.2i=-112.2i

0j-186.4j=-186.4j

arctan-186.4/112.2

=59 degrees S of E

to find magnitude

sqr((-112.2)^2+(-186.4)^2))

=217.6 km

I just don't know how to include the change of 15 km and 30 degrees midway through the problem. Can someone look over my work to see if thats correct and then do the whole problem?

Explanation / Answer

sum distance in the x
x = -290*cos(40 degrees) + x2 -15*sin(30 degrees) = 110

so x2 = 339.7 km

sum in the y
x = 290*sin(40 degrees) + y2 - 15*cos(30 degrees) = 0

y2=-173.4 km

so mag = sqrt(339.7^2 + 173.4^2)=381.4 km

direction = arctan(-173.4/339.7)= 27.04 south of east

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