A) A particle moves in the xy plane with a constant acceleration. At time zero,

Question

A) A particle moves in the xy plane with a constant acceleration. At time zero, the particle is at x=3m, y=8m, and has velocity vector vo= (6.5m/s)i+(-2.5m/s)j. The acceleration is given by vector a= (4m/s^2)i+(1.5m/s^2)j. What is the x component of velocity after 9s? Answer in units of m/s.

B) What is the y component of velocity after 9s? Answer in units of m/s.

C) What is the magnitude of the displacement from the orgin (x=0m,y=0m) after 9s? Answer in units of m.

I keep doing the problem and I'm geting x=223.5. Not sure what I'm doing wrong. If you could please fully work out the problem so I can look at my 4 attempts to compare to the correct answer.

Explanation / Answer

      At time t = 0 s, the displacement vector is                    r0 = (3.0 m)i^+(8.0 m)j^
      At time t = 0 s, the velocity vector is                    v0 = (6.5 m/s)i^+(-2.5 m/s)j^       At time t = 0 s, the acceleration vector is                   a0 = (4 m/s2)i^+(1.5 m/s2)j^ __________________________________________________________       The time t = 9 s       Using kinematic equation, we have                     v(t) = v0+at
      substitute the given data in above kinematic relation, we get            v(t = 9 s) = [(6.5 m/s)i^+(-2.5 m/s)j^]+[(4 m/s2)i^+(1.5 m/s2)j^][9 s]                           = [(6.5 m/s)i^+(-2.5 m/s)j^]+[(36 m/s)i^+(13.5 m/s)j^]                           = (42.5 m/s)i^+(11 m/s)j^ ___________________________________________________________ a)       Therefore, the x-component of the velocity after t = 9 s is                                 vx= 42.5 m/s ____________________________________________________________ ____________________________________________________________ b)       similarly, the y-component of the velocity after t = 9 s is                                 vy= 11 m/s ____________________________________________________________ ____________________________________________________________                                 vy= 11 m/s ____________________________________________________________ ____________________________________________________________ c)       Using kinematic equation, we have                 r = r0 + v0t+(1/2)at2                    = [(3.0 m)i^+(8.0 m)j^]+[(6.5 m/s)i^+(-2.5 m/s)j^][9 s]                       +(1/2)[(4 m/s2)i^+(1.5 m/s2)j^][9 s]2                    = [(3.0 m)i^+(8.0 m)j^]+[(58.5 m)i^+(-22.5 m/s)j^]                       +[(162 m)i^+(60.75 m)j^]                    = (223.5 m)i^+(46.25 m)j^                    = [(3.0 m)i^+(8.0 m)j^]+[(58.5 m)i^+(-22.5 m/s)j^]                       +[(162 m)i^+(60.75 m)j^]                    = (223.5 m)i^+(46.25 m)j^                       +[(162 m)i^+(60.75 m)j^]                    = (223.5 m)i^+(46.25 m)j^        Therefore, the magnitude of the displacement is                  | r | = (223.5 m)2+(46.25 m)2                       = 228.24 m                       = 228.24 m

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