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