Question 911969
A particle moves on a vertical line so that it's coordinate in feet at time (t) is s(t) = 4t^3 - 12t^2 +7   ,    t >= 0

a)
Find the Velocity function

b)
Find the Acceleration function

c)
When is the particle moving upward?

d)
When is the particle moving downward?

e)
What is the total distance that the particle travels in the time interval 0 <= t <= 3
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a)
Find the Velocity function


Differentiate the position function s(t) to get v(t) = s'(t) the velocity function


s(t) = 4t^3 - 12t^2 +7


s'(t) = 3*4t^2 - 2*12t^1 +0


s'(t) = 12t^2 - 24t


v(t) = 12t^2 - 24t


The velocity function is v(t) = 12t^2 - 24t

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b)
Find the Acceleration function


Differentiate the velocity function v(t) to get the acceleration function a(t) = v'(t). This is equivalent to finding the second derivative of s(t)


v(t) = 12t^2 - 24t


v'(t) = 2*12t^1 - 1*24t^0


v'(t) = 24t - 24*1


v'(t) = 24t - 24


a(t) = 24t - 24


The acceleration function is a(t) = 24t - 24

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c)
When is the particle moving upward?


The particle moves upward when v(t) > 0, essentially when the velocity is positive.


Graph v(t) = 12t^2 - 24t to get


{{{drawing(500,500,-10,10,-25,25,
  graph(500,500,-10,10,-25,25,0,12x^2-24x)
)}}}


We see that the velocity function is positive when {{{t < 0}}} or {{{t > 2}}}. The inequality {{{t < 0}}} does not make sense in this case because negative time is not possible. So only focus on {{{t > 2}}}


The particle is moving upward when {{{t > 2}}} (beyond 2 seconds)

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d)
When is the particle moving downward?


Use the same graph in part c) to find that the velocity is negative when {{{0<t<2}}}. This is when the particle is moving downward.

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e)
What is the total distance that the particle travels in the time interval 0 <= t <= 3


D = total distance the particle travels in the time interval 0 <= t <= 3


*[Tex \Large D = \int_{0}^{3}|v(t)|dt]


*[Tex \Large D = \int_{0}^{2}|v(t)|dt+\int_{2}^{3}|v(t)|dt]


*[Tex \Large D = -\int_{0}^{2}(12t^2 - 24t)dt+\int_{2}^{3}(12t^2 - 24t)dt]


*[Tex \Large D = -(4t^3 - 12t^2)|_{0}^{2}+(4t^3 - 12t^2)|_{2}^{3}]


*[Tex \Large D = -((4(2)^3 - 12(2)^2)-(4(0)^3 - 12(0)^2))+((4(3)^3 - 12(3)^2)-(4(2)^3 - 12(2)^2))]


*[Tex \Large D = 32]


The total distance traveled is 32 feet.