Aspaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v⃗ =9500i^m/s. it turns on its thruster engine and accelerates with a⃗ =(40i^−20k^)m/s2 for 35 min. part a what is the spaceship's position when the engine shuts off? give your answer as a vector measured in km.

 The spaceship's position when the engine shuts off =

Explanation:

  Initial location of spaceship = (600 i - 400 j + 200 k)*= (600 i - 400 j + 200 k)*

  Initial velocity = 9500 i m/s

  Acceleration = (40 i - 20 k)

  Time = 35 minute = 35 * 60 = 2100 seconds

 We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  Substituting

     So final position =

                              =

    The spaceship's position when the engine shuts off =

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