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The position of the particle is given by y(t) = (t³ − 4t² − 6t)m.

Mathematics

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Answer

22 m\text{22 m}

11 m/s\text{11 m/s}

Step-by-step solution

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Here's the solution for the object's displacement and average velocity:

The position of the particle is given by y(t)=(t3−4t2−6t)my(t) = (t^3 - 4t^2 - 6t)\text{m}.
We need to find the displacement and average velocity for the time interval from t=3st = 3\text{s} to t=5st = 5\text{s}.

  1. Calculate the position at t=3st = 3\text{s}.

    Substitute t=3t=3 into the position function:

    y(3)=(33−4(32)−6(3))my(3) = (3^3 - 4(3^2) - 6(3))\text{m} y(3)=(27−4(9)−18)my(3) = (27 - 4(9) - 18)\text{m} y(3)=(27−36−18)my(3) = (27 - 36 - 18)\text{m} y(3)=(−9−18)my(3) = (-9 - 18)\text{m} y(3)=−27my(3) = -27\text{m}
  2. Calculate the position at t=5st = 5\text{s}.

    Substitute t=5t=5 into the position function:

    y(5)=(53−4(52)−6(5))my(5) = (5^3 - 4(5^2) - 6(5))\text{m} y(5)=(125−4(25)−30)my(5) = (125 - 4(25) - 30)\text{m} y(5)=(125−100−30)my(5) = (125 - 100 - 30)\text{m} y(5)=(25−30)my(5) = (25 - 30)\text{m} y(5)=−5my(5) = -5\text{m}
  3. Calculate the displacement (Δy\Delta y).

    Displacement is the change in position, Δy=y(t2)−y(t1)\Delta y = y(t_2) - y(t_1).

    Δy=y(5)−y(3)\Delta y = y(5) - y(3) Δy=(−5m)−(−27m)\Delta y = (-5\text{m}) - (-27\text{m}) Δy=−5m+27m\Delta y = -5\text{m} + 27\text{m} Δy=22m\Delta y = 22\text{m}

    The object's displacement is 22 m\boxed{\text{22 m}}.

  4. Calculate the average velocity (vavgv_{avg}).

    Average velocity is the displacement divided by the time interval, vavg=ΔyΔtv_{avg} = \frac{\Delta y}{\Delta t}.
    The time interval is Δt=5s−3s=2s\Delta t = 5\text{s} - 3\text{s} = 2\text{s}.

    vavg=22m2sv_{avg} = \frac{22\text{m}}{2\text{s}} vavg=11m/sv_{avg} = 11\text{m/s}

    The object's average velocity is 11 m/s\boxed{\text{11 m/s}}.

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