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24 August, 04:42

Based on experimental observations, the acceleration of a particle is defined by the relation a = - (0.1 + sin x/b), where a and x are expressed in m/s2 and meters, respectively. Know that b = 0.98 m and that v = 1 m/s when x = 0.

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  1. 24 August, 06:14
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    a) v = + / - 0.515 m/s

    b) x = - 0.098164 m

    c) v = + / - 1.005 m/s

    Explanation:

    Given:

    The relationship for the acceleration is given as follows:

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

    Where, b = 0.98

    - IVP is 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/0.98))

    - Separate variables:

    v*dv = - (0.1 + sin (x/0.98)). dx

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

    0.5 * (v^2) = - (0.1*x - 0.98*cos (x/0.8)) - 0.98 + 0.5

    0.5*v^2 = 0.98*cos (x/0.98) - 0.1*x - 0.48

    - Evaluate at, x = - 1

    0.5*v^2 = 0.98 cos (-1/0.98) + 0.1 - 0.48

    v = sqrt (0.2651155)

    v = + / - 0.515 m/s

    - v = v_max when a = 0. Set the given expression to zero and solve for x:

    -0.1 = sin (x/0.98)

    x = - 0.98*0.1002

    x = - 0.098164 m

    - Hence now evaluate velocity through the derived expression:

    v^2 = 1.96 cos (-0.098164/0.98) - 0.96 - 0.2*-0.098164

    v = sqrt (1.0098)

    v = + / - 1.005 m/s
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