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The velocity function is v(t)= -t^2+3t-2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [0,5].

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18 answers

  1. well since they want displacement (vector) not distance (scalar) watch out for sign reversals. Like when is v = 0 ???
    t^2 - 3t +2 = 0
    (t -1)(t-2) = 0
    so do three intervals
    (a) t = 0 to t = 1
    (b) t = 1 to t = 2
    (c) t = 2 to t = 5

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  2. integral from p to q of (-t^2+3t-2 ) dt =-t^3/3 + 3t^2/2 - 2 t at q - at p
    = (-q^3+p^3) + (3/2)(q^2-p^2) -2(q-p)
    (a) from p=0 to q=1
    = -1 + 3/2 - 2 = -3 + 3/2 = -3/2
    (b) etc etc etc

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  3. I am still getting it wrong

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  4. since they want displacement, ignore sign changes.
    ∫[0,5] -t^2+3t-2 dt = -86/5

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  5. -86/5 isn't right

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  6. in that case there is a mistake in the problem or in the answer key. All it wants is just a straight integral. You can see that the particle moves
    left for 1 second
    right for 1 second
    left for 3 more seconds. It will wind up to the left of where it started.

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  7. displacement has to include sign changes
    if you start at home, walk north a mile, then south a mile
    your distance is TWO miles
    your displacement is ZERO miles

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  8. its asking for total distance traveled so the answer wont be negative

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  9. Yes, there must be sign changes

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  10. It asked for the NET distance, not the TOTAL distance
    Yes total is to the left

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  11. Huh? It said
    Find the displacement (net distance covered)
    not distance traveled.
    But, in that case, you want

    -∫[0,1] (-t^2+3t-2) dt + ∫[1,2] (-t^2+3t-2) dt - ∫[2,5] (-t^2+3t-2) dt
    = 5/6 + 1/6 + 27/2
    = 29/2

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  12. answer key said incorrect to 29/2

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  13. or that could be just ∫[0,5] |-t^2+3t-2| dt = 29/2
    but to do the actual evaluation, you need to split it up as above.

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  14. well, something is wrong. We've done it both ways.
    Maybe the answer key wants 14.5 instead of 29/2

    You're on your own now, buddy.

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  15. could it be you made a mistake?

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  16. I think the question is poorly worded and it is hard to say. Obviously the displacement is negative (you end up to the left), but the net distance traveled might mean the absolute value of the displacement. Whoever wrote the question does not do physics.

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  17. I think it is asking for the absolute value

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  18. I am not so sure about that, displacement, unlike distance, includes direction.

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