Two airplanes leave an airport at the same time. The first flies #(210 " km")/("hr")# in a direction of #290^@#. The second flies #(180 " km")/("hr")# in a direction of #185^@#. After 3 hr, how far apart are the planes?

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Answer 1 · 2018-04-30T04:50:59

#d_"apart" = 941.2" km"#

The magnitude of the first distance vector after 3 hours is:

#d_1 = (210 " km")/("hr")(3" hr") = 630" km"#

The magnitude of the second distance vector after 3 hours is:

#d_2= (185" km")/"hr"(3" hr") = 555" km"#

Using a variant of the Law of Cosines:

#d_"apart" = sqrt(d_1^2+d_2^2-2(d_1)(d_2)cos(290^@-185^@))#

#d_"apart" = sqrt((630" km")^2+(555" km")^2-2(630" km")(555" km")cos(105^@))#

#d_"apart" = 941.2" km"#