Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the ydirection of ay = 7.30 m/s2. The engines turn off after firing for 675 s, at which point the spacecraft has velocity components of vx = 3630 m/s and vy = 4276 m/s.What was the magnitude and direction of the spacecraft's initial velocity, before the engines were turned on?Express the direction as an angle measured counterclockwise from the x -axis.

Sagot :

Answer:

v₀ = 677.94 m / s ,   θ = 286º

Explanation:

We can solve this exercise using the kinematic expressions, let's work on each axis separately.

X axis

has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation

         vₓ = v₀ₓ + aₓ t

         v₀ₓ = vₓ - aₓ t

let's calculate

         v₀ₓ = 3630 - 5.10 675

         v₀ₓ = 187.5 m / s

Y Axis

        [tex]v_{y}[/tex] = v_{oy} - a_{y} t

         v_{oy} = v_{y} - a_{y} t

   

let's calculate

        v_{oy}  = 4276 - 7.30 675

         v_{oy} = -651.5 m / s

we can give the speed starts in two ways

a)   v₀ = (187.5 i ^ - 651.5 j ^) m / s

b) in the form of module and angle

Let's use the Pythagorean theorem

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

            v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]

            v₀ = 677.94 m / s

we use trigonometry

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

            θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }

            θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])

            θ = -73.94º

This angle measured from the positive side of the x-axis is

            θ‘ = 360 - 73.94

            θ = 286º

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.