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Find the position and velocity of a particle at t=1.94 s if the particle is initially moving east at a speed of 19.8 m/s and experiences an acceleration of magnitude 4.08 m/s^2, directed west.

Magnitude and direction of the position.
direction- East of its original position

Magnitude and direction of the velocity.
direction- towards the east

I have figured out the directions for both of them, but I have been unable to figure out the magnitudes, I have tried many equations but none seem to work correctly. An answer with a detailed explanation is greatly appreciated.


  • Physics - ,

    V = 19.8m/s, East.

    V = at=4.08m/s^2*1.94s = 7.92m/s,West.

    Vn = 19.8 - 7.92 = 11.88m/s, East.
    Vn = Net velocity.

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