Respuesta :
The vector addition and kinematics of uniform motion allows to find the result for where the starting point is:
- The ship left a point at a distance of 77,565 miles in a direction of 136.7º from the x-axis.
Given parameters
- First speed v₁ = 21.3 pH
- Angle θ₁ = 123º
- Time t₁ = 3.20 h
- Second speed v₂ = 18.2 mph
- Angle θ₂ = 190.0º
- Time t₂ = 1.10 h
To find
- Distance start point.
Velocity is a vector magnitude, therefore it has direction and module, to make the addition of these magnitudes vector algebra must be used.
There are two methods for adding vectors:
- Graphic. In this case, a vector is placed at the tip of the other and the resulting vector goes from the origin of the first to the tip of the last.
- Analytical. In this case, the vector is decomposed into a coordinate system, the resultants are found and the vector is composed.
In the attached we can see a diagram of the displacement of this ship. We use a Cartesian reference system and using trigonometric to decompose the velocities.
First speed
cos 123 = [tex]\frac{v_1_x}{v_1}[/tex]
sin 123 = [tex]\frac{v_1_y}{v_1}[/tex]
v₁ₓ = v₁ cos 123
[tex]v_{1y}[/tex] = v₁ sin 123
v₁ₓ = 21.1 cos 123 = -11.49 mph
[tex]v_{1y}[/tex] = 21.1 sin123 = 17.70 mph
Second speed
v₂ₓ = v₂ cos 190
[tex]v_{ 2y}[/tex] = v₂ sin190
v₂ₓ = 18.2 cos 190 = -17.92 mph
[tex]v_{2y}[/tex] = 18.2 sin190 = -3.16 mph
Now we can use kinematics of uniform motion.
v = [tex]\frac{d}{t}[/tex]
d = v t
Where v is the velocity, d the distance and t the time
Let's find every distance traveled.
For first speed
x-axis
x₁ = v₁ₓ t₁
x₁ = -11.49 3.20
x₁ = - 36,768 miles
y-axis
y₁ = [tex]v_{1y} \ t_1[/tex]
y₁ = 17.70 3.20
y₁ = 56.64 miles
Travel for 2nd speed
x₂ = v₂ₓ t₂
x₂ = -17.92 1.10
x₂ = -19,712 miles
y₂ = -3.16 1.10
y₂ = - 3,476 miles
Let's find the distance traveled on each axis
x-axis
[tex]x_{total}[/tex] = x₁ + x₂
[tex]x_{total}[/tex] = -36,768 - 19,712
[tex]x_{total}[/tex] = -56.48 miles
y-axis
[tex]y_{total}[/tex] = y₁ + y₂
[tex]y_{total}[/tex] = 56.64 - 3.476
[tex]y_{total}[/tex] = 53,164 miles
Let's construct the total displacement vector, for the module let's use the Pythagorean theorem
[tex]d_{total} = \sqrt{x^2 +y^2} \\d_{total} = \sqrt{56.48^2 +53.164^2}[/tex]
[tex]d_{total}[/tex] = 77,565 miles
Let's used trigonometry for the direction
[tex]tan \theta ' = \frac{y_{total} }{ x_{total} } \\\theta ' = tan^{-1} \frac{y_{total} }{ x_{total} }[/tex]
θ’= 43.3º
This angle is in the second quadrant to indicate the angle with respect to the positive side of the x-axis in a counterclockwise direction.
θ = 180 - θ'
θ = 180 - 43.3
θ = 136.7º
In conclusion using the addition vectors and kinematics of uniform motion we can find the result for where the starting point is:
- The ship left a point at a distance of 77,565 miles in a direction of 136.7º from the x-axis.
Learn more here: brainly.com/question/15074838
