A radar station sends a signal to a ship which is located a distance 13.9 kilometers from the station at bearing 136° clockwise from the north. At the same moment, a helicopter is at a horizontal range of 19.6 kilometers, at bearing 152° clockwise from the north, with an elevation of 2.57 kilometers. Let east be the î direction, north be the ĵ direction, and up be the direction.
(A) What is the displacement vector (in km) from the helicopter to the ship? (Express your answer in vector form. Do not include units in your answer.)
(B) What is the distance (in km) between the helicopter and ship?
(C) The ship begins to sink at a rate of 5.40 m/s. Write the position vector (in km) of the ship relative to the helicopter as a function of time as the ship sinks. Assume that the helicopter remains hovering at its initial position and that the sinking rate remains the same even after the ship sinks under the surface. (Use the following as necessary: t. Do not include units in your answer.)
(D) Evaluate the position of the ship relative to the helicopter (in km) after 1.90 h. (Express your answer in vector form. Do not include units in your answer.)

Respuesta :

Answer:

A)  d = 7.9267 km,  θ = 157.19

D) d = 35.17 km

Explanation:

For this exercise the easiest way is to decompose the vectors

Ship

X axis  (North)

The angle measured from the positive side of the x axis (unit vector i^ ) counterclockwise

           θ = 136- 136 = 224

           

           cos θ = x₁ / 13.9  

           x₁ = 13.9  cosc 224

           x₁ = - 9.9988 103 m

z axis

           sin 224 = z₁ / 13.9

           z₁ = 13.9 sin224

           z₁ = -9.6558 m

Axis y

            y₁ = 0

 

Helicopter

x-axis (North)

         θ = 360 - 152

         θ = 208

          cos θ = x₂ / 19.6

          x₂ = 19.6 cos 208

          x₂ = -17.3058 km

z axis

           sin 204 = z₂ / 19.6

           z₂ = 19.6 sin 204

           z₂ = -7.9720 km

Axis y

           y₂ = 2.57 km

It's a bit strange the way the axes give, but having the decomposition of the vectors we can answer the questions

A) They ask the displacement vector, let's start looking for the distance

            d = √ [(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²]

             

we substitute

            d =√[(17.3058 - 9.9988)² + (2.57 - 0)² + (7.9720 -9.6558)²]

            d = √ [7.307² + 2.57² + 1.6838²]

            d = 7.9267 km

the angle of this vector is with respect to North (x-axis)

            cos θ = -7.307 / 7.9267  

            θ = cos⁻¹ (-0.9218)

            θ = 157.19

Y-axis elevation angle

           cos θ'= 2.57 / 7.9267

           θ’= cos⁻¹ (0.3242)

           θ’= 71.09

 

b) d = 7.9267 km

c) at this time the ship is sinking at constant speed, the sinking direction is the same vertical direction of the helicopter (y-axis)

          the distance is

               y₁ = v₁ t

let's reduce speed to km /h

              v₁ = 5.4 m / s (1km/1000m ) (3600s/1h)= 19.44 km /h

              y₁ = 19.44 t

the vector positions is

             d = √ [(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²]

            d = √ [7.307² + (2.57 -19.44 t)² + 1.6838²]

in vertical angle is

          cos θ'= (2.57 -19.44t) / d

At an angle to the x axis is

         cos θ = -7.307 / d

d) let's evaluate the position for t = 1.90 h    

distance

       d = √ [7.307² + (2.57- 19.44 1.9)² + 1.6838²]

       d = √ [7.307² + (-34.366)² + 1.6838²]

       d = 35.17 km