Question

During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal

Answer 1

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 = $\frac{\Delta d}{t}$

              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) = $\frac{x_1}{\Delta d_1}$

          sin 145 = $\frac{y_1}{\Delta d1}$

          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) = $\frac{x_2}{\Delta d_2}$

          sin (120) = $\frac{y_2}{\Delta d_2}$

          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 = $\sqrt{R_x^2 +R_y^2}$

             R = $\sqrt{70.98^2 + 62.08^2}$   10³

             R = 94.30 10³ m

We use trigonometry for the angle

             tan θ ’= $\frac{R_y}{R_x}$

             θ '= tan⁻¹ $\frac{R_y}{R_x}$

             θ '= tan⁻¹ $\frac{62.08}{70.98}$

             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 = $\frac{\Delta R}{t_3}$  

             v = $\frac{94.30}{120} \ 10^3$

              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