Draw additional masses to show how the block can be made to move toward point A. Be sure to draw the string and label the mass o
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The kinematics of the uniform motion and the addition of vectors allow finding the results are:
- The Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º
- The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East
Vectors are quantities that have modulus and direction, so they must be added using vector algebra.
A simple method to perform this addition in the algebraic method which has several parts:
- Vectors are decomposed into a coordinate system
- The components are added
- The resulting vector is constructed
Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case
v =
Δd = v t
Where v is the average velocity, Δd the displacement and t the time
We look for the first distance traveled at speed v₁ = 600 m / s for a time
t₁ = 2 min = 120 s
Δd₁ = v₁ t₁
Δd₁ = 600 120
Δd₁ = 72 10³ m
Now we look for the second distance traveled for the velocity v₂ = 400 m/s
time t₂ = 1 min = 60 s
Δd₂ = v₂ t₂
Δd₂ = 400 60
Δd₂ = 24 103 m
In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,
We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)
Let's use trigonometry for each distance
Route 1
cos (180 -35) =
sin 145 =
x₁ = Δd₁ cos 125
y₁ = Δd₁ sin 125
x₁ = 72 103 are 145 = -58.98 103 m
y₁ = 72 103 sin 155 = 41.30 10³ m
Route 2
cos (90+ 30) =
sin (120) =
x₂ = Δd₂ cos 120
y₂ = Δd₂ sin 120
x₂ = 24 103 cos 120 = -12 10³ m
y₂ = 24 103 sin 120 = 20,78 10³ m
The component of the resultant vector are
Rₓ = x₁ + x₂
R_y = y₁ + y₂
Rx = - (58.98 + 12) 10³ = -70.98 10³ m
Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m
We construct the resulting vector
Let's use the Pythagoras' Theorem for the module
R =
R = 10³
R = 94.30 10³ m
We use trigonometry for the angle
tan θ ’=
θ '= tan⁻¹
θ '= tan⁻¹
θ ’= 41.2º
Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction
θ = 180 - θ'
θ = 180 -41.2
θ = 138.8º
Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.
The average speed of the trip should be
v =
v =
v = 785.9 m / s
in the opposite direction, that is, the angle must be
41.2º to the South of the East
In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:
- To find the initial Barry trajectory is 94.30 10³ m with n angles of 138.8º
- The return trajectory and speed is v = 785.9 m / s, with an angle of 41.2º to the South of the East
Learn more here: brainly.com/question/15074838
