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28 October, 23:08

Proposed Kinematic Exercise I

Consider a body moving on a straight path with a speed given by v = (3t^2-6t) m/s, where t is in seconds. If the body is at x=4m when t=0s, determine (a) its position at t=4s and (b) its acceleration at t=2s. (c) What is the direction of body movement between t=0s and t=4s? Justify your answer.

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  1. 29 October, 01:58
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    (a) 20 m

    (b) 6 m/s²

    (c) Between t=0 and t=2, the body moves to the left.

    Between t=2 and t=4, the body moves to the right.

    Explanation:

    v = 3t² - 6t

    x (0) = 4

    (a) Position is the integral of velocity.

    x = ∫ v dt

    x = ∫ (3t² - 6t) dt

    x = t³ - 3t² + C

    Use initial condition to find value of C.

    4 = 0³ - 3 (0) ² + C

    4 = C

    x = t³ - 3t² + 4

    Find position at t = 4.

    x = 4³ - 3 (4) ² + 4

    x = 20

    (b) Acceleration is the derivative of velocity.

    a = dv/dt

    a = 6t - 6

    Find acceleration at t = 2.

    a = 6 (2) - 6

    a = 6

    (c) v = 3t² - 6t

    v = 3t (t - 2)

    The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.

    When 0 < t < 2, v < 0.

    When t > 2, v > 0.
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