# Answer to Question #14372 in Mechanics | Relativity for Taylor

Question #14372

A radar station locates a sinking ship at range 17.5 km and bearing 136° clockwise from north. From the same station, a rescue plane is at horizontal range 19.6 km, 169° clockwise from north, with elevation 2.40 km.

(a) Write the displacement vector from plane to ship, letting represent east, north, and up.

(b) How far apart are the plane and ship?

(a) Write the displacement vector from plane to ship, letting represent east, north, and up.

(b) How far apart are the plane and ship?

Expert's answer

(a) Write the displacement vector from plane to ship, letting

represent

east, north, and up.

From station to ship:

East: 17.5 * sin(136°) =

12.156 km

North: 17.5 * cos(136°) = -12.588 km

Up: 0 km

From station to

plane:

East: 19.6 * sin(169°) = 3.740 km

North: 19.6 * cos(169°) = -19.240

km

Up: 2.4 km

From plane to ship:

East: 17.5 * sin(136°) = 12.156

km

North: 17.5 * cos(136°) = -12.588 km

Up: 0 km

From station to

plane:

East: 19.6 * sin(169°) = - 8.416 km

North: 19.6 * cos(169°) = -

6.652 km

Up: 2.4 km

(b) How far apart are the plane and ship?

At

horizontal range: Lh = sqrt( (- 8.416)^2 + (-6.652)^2 ) = 10.727 km

At all L=

sqrt( Lh^2 + 2.4^2 ) = 10.993 km

represent

east, north, and up.

From station to ship:

East: 17.5 * sin(136°) =

12.156 km

North: 17.5 * cos(136°) = -12.588 km

Up: 0 km

From station to

plane:

East: 19.6 * sin(169°) = 3.740 km

North: 19.6 * cos(169°) = -19.240

km

Up: 2.4 km

From plane to ship:

East: 17.5 * sin(136°) = 12.156

km

North: 17.5 * cos(136°) = -12.588 km

Up: 0 km

From station to

plane:

East: 19.6 * sin(169°) = - 8.416 km

North: 19.6 * cos(169°) = -

6.652 km

Up: 2.4 km

(b) How far apart are the plane and ship?

At

horizontal range: Lh = sqrt( (- 8.416)^2 + (-6.652)^2 ) = 10.727 km

At all L=

sqrt( Lh^2 + 2.4^2 ) = 10.993 km

## Comments

## Leave a comment