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26 March, 09:13

Based on experimental observations, the acceleration of a particle is defined by the relationa = - (0.1 + sin (x/b)), where a and x are expressed in m/s^2 and meters, respectively. Knowing that b = 0.8 m and that v = 1 m/s when x = 0, determine (a) the velocity of the particle when x = - 1 m, (b) the position where the velocity is maximum, (c) the maximum velocity.

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  1. 26 March, 09:26
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    a) v = + / - 0.323 m/s

    b) x = - 0.080134 m

    c) v = + / - 1.004 m/s

    Explanation:

    Given:

    a = - (0.1 + sin (x/b))

    b = 0.8

    v = 1 m/s @ x = 0

    Find:

    (a) the velocity of the particle when x = - 1 m

    (b) the position where the velocity is maximum

    (c) the maximum velocity.

    Solution:

    - We will compute the velocity by integrating a by dt.

    a = v*dv / dx = - (0.1 + sin (x/0.8))

    - Separate variables:

    v*dv = - (0.1 + sin (x/0.8)). dx

    -Integrate from v = 1 m/s @ x = 0:

    0.5 (v^2) = - (0.1x - 0.8cos (x/0.8)) - 0.8 + 0.5

    0.5v^2 = 0.8cos (x/0.8) - 0.1x - 0.3

    - Evaluate @ x = - 1

    0.5v^2 = 0.8 cos (-1/0.8) + 0.1 - 0.3

    v = sqrt (0.104516)

    v = + / - 0.323 m/s

    - v = v_max when a = 0:

    -0.1 = sin (x/0.8)

    x = - 0.8*0.1002

    x = - 0.080134 m

    - Hence,

    v^2 = 1.6 cos (-0.080134/0.8) - 0.6 - 0.2*-0.080134

    v = sqrt (0.504)

    v = + / - 1.004 m/s
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