Post a New Question


posted by .

A space vehicle is coasting at a constant velocity of 24.7 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 0.228 m/s2 in the +x direction. After 42.3 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.

  • physics -

    V(hor) = 0.228m/s^2 * 42.3s = 9.64m/s.

    a. V^2 + Y^2=(9.64)^2 + (24.7)^2 = 703,
    V = sqrt(703) = 26.5m/s = Magnitude.

    b. tanA = Y/X = 24.7 / 9.64 = 2.5622,
    A = 68.7Deg CCW = 90 - 68.7 = 21.3Deg
    East of +Y = Direction.

  • physics -

    CORRECTION: a. X^2 + Y^2 =

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question