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anonymous

  • one year ago

At t = 0, a particle is located at x = 25 m and has a velocity of 15 m/s in the positive x direction. The acceleration of the particle varies with time as shown in the diagram. What is the position of the particle at t = 5.0 s? The acceleration-time graph looks like a straight line and the slope is negative and it makes a triangle with base of 6 units and a height of 6 units.

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  1. IrishBoy123
    • one year ago
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    |dw:1444427089918:dw|

  2. IrishBoy123
    • one year ago
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    if that drawing is correct, you have \(a(t) = 6-t\) you know that velocity \(v(t) = \int a(t) \; dt \; [ [A]]\) you know that displacement/position is given by \(x(t) = \int \; v(t) \; dt\; [ [B] ]\) you also have boundary conditions: \( x(0) = 25 m, v(0) = +15 m/s\) integrate [A] then [B], and then use the conditions to solve for the constants then solve for x(5).

  3. anonymous
    • one year ago
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    Sorry for taking so long to replay but the site was down for me for some reason. That worked perfectly i got the equation from the graph and integrated it twice like you said and i got the correct answer, so thank you!

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