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A space vehicle is coasting at a constant velocity of 24.1 m/s in the +y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.240 m/s2 in the +x direction. After 58.7 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the +y direction.

Respuesta :

Answer:

The speed is 26.5 m / s and 29.9º

Explanation:

This is a kinematic problem, let's look at the speed in the x direction, after the RCS shutdown time

      Vfx = Vox + a t

Initially there was no velocity on the x-axis, which implies that Vox = 0

     Vf = at

     Vf = 0.240 58.7

     Vfx = 13,872 m / s

The magnitude of the velocity, we can find it using the Pythagorean theorem

     

     V² = Vy² + Vx²

     V = √ (24.1² + 13,872²)

     V = 26.48 m / s

The direction can be found by trigonometry, as we are asked for the angle measured with respect to the axis and

     Tan θ = Vx / Vy

     θ = tan -1 (13,872 / 24.1)

     θ = 29.9º

The speed is 26.5 m / s and 29.9º

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