Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

1. On a spacecraft two engines fire for a time of 565 s. One gives the craft an acceleration in the x direction of ax = 5.10 m/s^2, while the other produces an acceleration in the y direction of ay = 7.30 m/s^2. At the end of the firing period, the craft has velocity components of vx = 3775 m/s and vy = 4816 m/s. Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the +x axis.

Sagot :

The kinematics allows to find the result for the initial speed of the rocket is

         v₀ = 2.23 m / s with a direction of θ = 54.1º

Kinematics studies the movement of bodies, finding relationships between position, velocity and acceleration.

The reference system is a coordinate system with respect to which all measurements must be made, in this case they indicate the accelerations and veloicties with respect to an x ​​and y coordinate system,

The accelerations are aₓ = 5.10 m/s² and [tex]a_y[/tex] = 7.30 m/s², it also indicates the time that the motors are on t = 565 s, when the acceleration is finished the final speeds are vₓ = 3775 m/s and [tex]v_y[/tex] = 4816 m/s.

Let's use the relationship

             vₓ = [tex]v_{ox}[/tex] + aₓ t

             [tex]v_{ox} = \frac{v_x}{a \ t}[/tex]  

Let's calculate

             v₀ₓ = [tex]\frac{3775}{5.1 \ 565}[/tex]

             v₀ₓ = 1.31 m / s

We perform the same calculation for the y axis

            [tex]v_y = v_{oy} + a_y t \\v_{oy} = \frac{v_y}{a_y t}[/tex]

            [tex]v_{oy} = \frac{4816}{ 7.3 \ 565 }[/tex]  

            [tex]v_{oy}[/tex] = 1.81 m / s

To find the velocity vector in the form of a module and direction we can use the Pythagoras' Theorem

            v₀ = [tex]\sqrt{v_{ox}^2 + v_{oy}^2}[/tex]

            v₀ = [tex]\sqrt{1.31^2 + 1.81^2}[/tex]

            v₀ = 2.23 m / s

To find the direction we use trigonometry

           tan θ = [tex]\frac{v_{oy}}{v_{ox}}[/tex]

           θ = tan⁻¹  [tex]\frac{v_{oy}}{v_{ox}}[/tex]

           θ = tan⁻¹  [tex]\frac{1.81}{1.31}[/tex]  

           θ = 54.1º

In conclusion with kinematics we can find the result for the initial speed of the rocket is

         v₀ = 2.23 m / s with a direction of  θ = 54.1º

Learn more here:  brainly.com/question/13202575

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.