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23 April, 19:29

The position of a particle moving along a coordinate line is s = √63 + 6t, with s in meters and t in seconds. Find the rate of change of the particle's position at t = 3 sec.

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Answers (2)
  1. 23 April, 22:48
    0
    0.33 m/s

    Step-by-step explanation:

    Given,

    s = √ (63+6t) ... Equation 1

    s' = ds/dt

    Where s' = rate of change of the particles position.

    Differentiating equation 1,

    s = (63+6t) ¹/²

    s' = 6*1/2 (63+6t) ⁻¹/²

    s' = 3 (63+6t) ⁻¹/²

    s' = 3/√ (63+6t) ... Equation 2

    At t = 3 s,

    Substitute the value of t into equation 2

    s' = 3/√ (63+6*3)

    s' = 3/√ (63+18)

    s' = 3/√ (81)

    s' = 3/9

    s' = 0.33 m/s.

    Hence the rate of change of the particles position = 0.33 m/s
  2. 23 April, 23:24
    0
    ds/dt = 0.33 m/s

    Therefore, the rate of change of the particle's position at t = 3 sec is 0.33 m/s

    Step-by-step explanation:

    Given;

    The position function of the particle.

    s (t) = √ (63+6t)

    The rate of change of the particle's position = ds/dt = s (t) '

    Using function of function rule.

    Let u = 63+6t

    s = √u

    ds/dt = du/dt * ds/du

    du/dt = 6

    ds/du = 0.5u^ (-0.5) = 0.5/u^ (0.5) = 0.5 / (63+6t) ^ (0.5)

    ds/dt = 6 * 0.5 / (63+6t) ^ (0.5)

    ds/dt = 3 / (63+6t) ^ (0.5)

    At t = 3sec

    ds/dt = 3 / (63+6 (3)) ^ (0.5) = 3/9

    ds/dt = 0.33 m/s

    Therefore, the rate of change of the particle's position at t = 3 sec is 0.33 m/s
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