Calculus

A particle moves along the x-axis so that at time "t" its position is given by s(t)=(t+3)(t-1)^3 ,t>0
For what values of "t" is the velocity of the particle increasing?

a) 0<t<1
b) t>1
c) t>0
d) the velocity is never increasing

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  1. You know that the velocity of s(t) is the first derivative of s(t)=(t+3)(t-1)^3 for t > 0
    s ' (t) = (t+3)(3)(t-1)^2 + (t-1)^3
    = (t-1)^2 (3t + 9 + t-1)
    = (t-1)^2 (4t + 8)

    you also know that if the function increases, (t-1)^2 (4t + 8) must be positive
    so the factor (t-1)^2 is always positive except when t = 1
    and since t > 0, 4t+8 will be always positive

    so for t > 0, s(t) will be increasing for all values of t, except t = 1
    check: https://www.wolframalpha.com/input/?i=plot+s(t)%3D(t%2B3)(t-1)%5E3

    your choices:
    a) 0<t<1 ----- correct but does not include all values of t
    b) t>1 -------- correct but does not include all values of t
    c) t>0 -------- true except when t = 1
    when t = 1, the function is "stationary"
    d) clearly false

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  2. s = t^4 - 6 t^2 + 8 t - 3 (check my multipication)
    v = ds/dt = 4 t^3 - 12 t + 8
    a = d^2 s/dt^2 = 12 t^2 -12 = 12 (t^2-1)
    where is that acceleration positive?
    where t^2 >1
    but t is only positive so
    t > 1

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  3. Okay I understand how you got the answer, thank you! But which answer choice would I pick? Thanks in advance.

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  4. Actually I get it, thanks again

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