Calculus

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a particle starts at time t = 0 and moves along the x axis so that its position at any time t>= 0 is given by x(t) = ((t-1)^3)(2t-3)
a.find the velocity of the particle at any time t>= 0
b. for what values of t is the velocity of the particle negative?
c. find the value of t when the particle is moving and the acceleration is zero. explain your answer choice

  • Calculus -

    a) find x'(t) using the product rule, that will be your velocity

    b) set x'(t) from above < 0 and solve

    c) acceleration is the derivative of velocity, so differentiate x'(t) again, and set it equal to zero

  • Calculus -

    a) Take the deriviative of x(t)
    b) Set the derivative = 0 and solve the equation for t
    c) Take the derivative of v(t) and set that equal to zero. If there is more than one answer, take the one for which v is not zero.

    We will be glad to critique your work.

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