Calculus

The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = 3t − 7, 0 ≤ t ≤ 3
(a) Find the displacement.
-7.5 m
(b) Find the distance traveled by the particle during the given time interval.

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  1. negative v at start, positive at 3 so reverses direction during the 3 seconds at t = 7/3
    v = 3 t - 7
    for vector displacement
    x = (3/2)t^2 - 7 t + c
    at t = 3, x = 27/2 - 21 + c
    at t = 0, x = c
    so
    displacement = -15/2 = -7.5 agree
    then
    for distance
    integrate from t = 0 to t = 7/3
    x = (3/2)t^2 - 7 t + c = (3/2)(49/9) - 49/3 = -49 /6 , negative motion
    then integrate from t = 7/3 to t = 3
    at t= 3
    x = 27/2 - 21 + c
    at t = 7/3
    x = (3/2)(49/9) -49/3+ c = -49/6 +c
    motion = 27/2 -21+49/6 = + 2/3 positive motion
    total distance moved = 49/6 + 4/6 = 53/6

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