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8 June, 20:17

The motion of a particle is defined by the relation x = t3 - 6t2 + 9t + 3, where x and t are expressed in feet and seconds, respectively. Determine When the velocity is zero The position, acceleration and total distance traveled when t = 5 sec

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  1. 8 June, 22:52
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    a. t=3secs and t=1sec

    position is - 7ft, acceleration is 18fts⁻², total distance travelled is 120ft

    Explanation:

    the displacement is define as

    x=t³-6t²+9t+3·

    since we are giving the position as a function of time, the velocity is the derivative of the position,

    v=dx/dt

    v=d (t³-6t²+9t+3) / dt

    recall for y=axⁿ the derivative

    dy/dx=a*nxⁿ⁻¹ and the derivative of a constant is zero

    hence

    V=3t²-12t+9

    for V=0,

    equivalent to t²-4t+3

    solving the quadratic equation, we arrive at

    (t-3) (t-1) = 0

    either t=3 or t=1

    hence, at 3secs and 1sec the velocity is zero.

    To determine the position at t=5, we substitute t=5 into

    t³-6t²+9t+3

    (5) ³-6 (5) ²+9 (5) + 3

    125-180+45+3

    -7ft

    The position at t=5 is - 7ft

    To determine the acceleration, we differentiate the velocity

    a=dv/dt

    a=d (3t²-12t+9) / dt

    a=6t-12

    at t=5

    a=6 (5) - 12

    a=18fts⁻²

    Next we determine the distance covered at t=5

    velocity = total distance travelled/total time taken

    velocity=3t²-12t+9

    V=3 (25) - 12 (5) + 9

    V=24ft/s

    Hence total distance travelled in t=5 is

    24*5=120ft
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