The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, t

Question

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)

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Phúc Điền 4 months 2021-09-05T09:40:49+00:00 1 Answers 2 views 0

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    2021-09-05T09:42:43+00:00

    Answer:

    v_max = (1/6)e^-1 a

    Explanation:

    You have the following equation for the instantaneous speed of a particle:

    v(t)=ate^{-6t}   (1)

    To find the expression for the maximum speed in terms of the acceleration “a”, you first derivative v(t) respect to time t:

    \frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

    where you have use the derivative of a product.

    Next, you equal the expression (2) to zero in order to calculate t:

    a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

    For t = 1/6 you obtain the maximum speed.

    Then, you replace that value of t in the expression (1):

    v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

    hence, the maximum speed is v_max = ((1/6)e^-1)a

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