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austinL

  • 3 years ago

At a certain time a particle had a speed of 80 m/s in the positive x direction, and 7.0 s later its speed was 90 m/s in the opposite direction. What was the average acceleration of the particle during this 7.0 s interval? I can think of two answers that I could get: 24.29m/s^2 1.43 m/s^2 Any help would be appreciated.

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  1. MarcLeclair
    • 3 years ago
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    I got 24.29 too but I haven't done these kind of problems in a long time. I just looked at the change of speed ( so from 80 to 0 and 0 to 90 considering it is in the opposite direction). So I put the acceleration as 24.28. ^^ ( well a = speed / delta t

  2. MarcLeclair
    • 3 years ago
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    or if you want to spend the time on it, draw the graph, find the rule of the velocity graph ( should be a linear one ax+b) and integrate it ^^

  3. austinL
    • 3 years ago
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    Ok, thanks! I just wanted another set of eyes to look at it.

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