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Average Acceleration?

Average Acceleration?

  • At a certain time a particle had a speed of 34 m/s in the positive x direction, and 8.1 s later its speed was 71 m/s in the opposite direction. What was the average acceleration of the particle during this 8.1 s interval? I know the average acceleration is = change of velocity/change of time so ... (71-34)/8.1= 4.57 m/s^2 but it says my answer is incorrect... please help... i feel dumb for having to ask this question

  • Answer:

    If the particle's direction of travel didn't reverse, then your calculation would be correct. Since the direction reversed, one or the other velocities is negative with respect to the other. The calculation is: [34 - (-71)] / 8.1 = 105 / 8.1 = 13.0 m/s^2

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Maybe need to see it as electrónic pulse and divide it in 4stages of frecuency and considere the pause of acelerecion betwen cicles more or lees acording of posición of the partcle at that moment probrably your aproch have to change

ALANVISION

use the relation a = v - u / t = 71 - (-34) / 8.1 = 71 +34 / 8.1 = 12.96 m/sec^2.

knr

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