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    Solid mechanics with a positive reference

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    1. A 2-lb ball is thrown upward at 30ft/s from an origin 5-ft above the earth's surface. Using for your positive reference a vertical Y axis with its origin at 5ft, DETERMINE FOR THE BALL:

    a) The maximum height in ft reached
    b) The elapsed time of flight in seconds until its return to the origin
    c) Its acceleration in ft/s2 at the highest point
    d) Its velocity vector at its return to the origin

    2. The velocity of a 0.1kg particle along a horizontal X axis is given by:
    v(t)=(8t2 -20)m/s, and at t=20s, v(20)=60m. DETERMINE FOR t=8s:

    a) The position of the particle in ft
    b) The velocity of the particle in ft/s
    c) The acceleration of the particle in ft/s2

    3. The acceleration of an object in rectilinear x-direction motion is given by:
    a(v)=0.4-0.0002v2 ft/s. If v=0 when x=0. DETERMINE FOR THE OBJECT:

    e) Its velocity in ft/s when x=1000ft
    f) Its acceleration in ft/s2 at x=1000ft
    g) Its maximum velocity in ft/s

    © BrainMass Inc. brainmass.com December 24, 2021, 7:04 pm ad1c9bdddf
    https://brainmass.com/physics/velocity/solid-mechanics-positive-reference-161165

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    SOLUTION This solution is FREE courtesy of BrainMass!

    Please see the attached file.

    The velocity , where s is the displacement of the object,
    And the acceleration

    On the other hand, we have
    The velocity is
    The displacement

    1. A 2-lb ball is thrown upward at 30ft/s from an origin 5-ft above the earth's surface. Using for your positive reference a vertical Y axis with its origin at 5ft, DETERMINE FOR THE BALL:

    a) The maximum height in ft reached
    b) The elapsed time of flight in seconds until its return to the origin
    c) Its acceleration in ft/s2 at the highest point
    d) Its velocity vector at its return to the origin

    The ball is subject to the gravitational acceleration g all the time.
    When the positive direction is upward, the acceleration a of the ball is
    a = - g = - 32 ft/s2.

    So the velocity is
    , where c1 is the integration constant, which is determined by the initial velcotiy.
    , so
    The velocity function is then
    ft/s.

    The height is

    When t = 0, h = 0, so
    The height function is

    (a) the maximum height:
    First find the critical point
    s
    Now according to the second derivative test, , so at , the height function h(t) has a maximum.
    The maximum height is
    ft.
    The maximum height is 14.0625 ft from the defined origin.
    It is 14.0625 + 5 = 19.0625 ft high from the earth's surface.

    (b) when the ball returns back to the origin, the height h(t) = 0

    Apparently, it takes 1.875 seconds for the ball to return back to the origin.

    (c) the ball has a constant acceleration of -32 ft/s2. so the acceleration is -32 ft/s2 at the highest point.

    (d) when t = 1.875 s, the ball returns back to the origin, so
    ft/s.
    The velocity is -30 ft/s when returning back to the origin.

    2. The velocity of a 0.1kg particle along a horizontal X axis is given by:
    v(t)=(8t2 -20)m/s, and at t=20s, v(20)=60m. DETERMINE FOR t=8s:
    Apparently, when t = 20 s, the velocity v(20) = 3180 m/s. So I think the red-highlighted parts should be t = 20 s, d(20) = 60 m, where d is the position of the particle.
    a) The position of the particle in ft

    When t = 20, d(20) = 60, so

    The position is
    meter
    So when t = 8 s,
    meter.
    The position of the particle is -19668 meters at t = 8.

    b) The velocity of the particle in ft/s
    When t = 8 s,
    m/s
    c) The acceleration of the particle in ft/s2
    The acceleration is
    , unit: m/s2
    When t = 8 s
    m/s2.

    All units are in meters, you can convert to ft using
    1 m =
    For example, the position of the particle is then
    The velocity is
    The acceleration is

    3. The acceleration of an object in rectilinear x-direction motion is given by:
    a(v)=0.4-0.0002v2 ft/s. If v=0 when x=0. DETERMINE FOR THE OBJECT:

    e) Its velocity in ft/s when x=1000ft
    f) Its acceleration in ft/s2 at x=1000ft
    g) Its maximum velocity in ft/s

    The acceleration
    So we have

    Integrate on both sides

    When x = 0, v = 0, so

    Take the exponential on both sides, we have

    (a) when x = 1000 ft,

    Then the velocity is
    or -25.68 ft/s

    (b) the acceleration is

    When x = 1000 ft,

    (c) The velocity function is

    We know that the exponential function is always positive, so when it is at its minimum, is at is maximum, that is, is at its maximum.

    So the maximum of v2 is

    That maximum speed is 44.72 ft/s.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:04 pm ad1c9bdddf>
    https://brainmass.com/physics/velocity/solid-mechanics-positive-reference-161165

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