A particle moves along a horizontal line so that its position at time t, t ≥ 0, is given by s(t) = 40 + te^−t/20. Find the min

Question

A particle moves along a horizontal line so that its position at time t, t ≥ 0, is given by
s(t) = 40 + te^−t/20.
Find the minimum velocity of the particle for 0 ≤ t ≤ 100.

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Thành Công 3 days 2021-07-22T17:24:54+00:00 1 Answers 3 views 0

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    2021-07-22T17:26:11+00:00

    Answer:

    The minimum velocity of the particle  = -e^{-2 } units

    Step-by-step explanation:

    Given – A particle moves along a horizontal line so that its position at time t,

    t ≥ 0, is given by  s(t) = 40 + te^−t/20.

    To find – Find the minimum velocity of the particle for 0 ≤ t ≤ 100.

    Proof –

    Velocity, v(t)  = \frac{d}{dt}(40 + te^{-\frac{t}{20} } )

    Now,

    \frac{d}{dt}(40 + te^{-\frac{t}{20} } ) =  \frac{d}{dt}(40 ) + \frac{d}{dt}(te^{-\frac{t}{20} } )

                           = 0 + t\frac{d}{dt}(e^{-\frac{t}{20} } ) + e^{-\frac{t}{20} }\frac{d}{dt}(t )

                           = t(-\frac{1}{20} )e^{-\frac{t}{20} } +  e^{-\frac{t}{20} }

    ⇒v(t) = -\frac{t}{20}e^{-\frac{t}{20} } + e^{-\frac{t}{20} }

    Now,

    For minimum velocity, Put \frac{d}{dt}(v(t)) = 0

    Now,

    \frac{d}{dt}[v(t)]  = \frac{d}{dt} [ -\frac{t}{20}e^{-\frac{t}{20} } + e^{-\frac{t}{20} } ]

               = -\frac{2}{20} e^{-\frac{t}{20} } + \frac{t}{400} e^{-\frac{t}{20} }

    Now,

    Put \frac{d}{dt}(v(t)) = 0, we get

    -\frac{2}{20} = - \frac{t}{400}

    ⇒t = 40

    Now,

    Check that the point is minimum or maximum

    Calculate  \frac{d^{2} }{dt^{2} } [v(t)]

    Now,

    \frac{d^{2} }{dt^{2} } [v(t)] = \frac{d}{dt} [ -\frac{2}{20} e^{-\frac{t}{20} } + \frac{t}{400} e^{-\frac{t}{20} }]

                = \frac{1}{400}e^{- \frac{t}{20} } [ 3 - \frac{t}{20}]

    \frac{d^{2} }{dt^{2} } [v(t)]  = \frac{1}{400}e^{- \frac{t}{20} } [ 3 - \frac{t}{20}]  > 0

    ∴ we get

    t = 40 is point of minimum

    So,

    The minimum velocity be

    v(40) = -\frac{40}{20}e^{-\frac{40}{20} } + e^{-\frac{40}{20} }

          = -2e^{-2 } + e^{-2 }

          = -e^{-2 }

    ⇒v(40) = -e^{-2 } units

    ∴ we get

    The minimum velocity of the particle  = -e^{-2 } units

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