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

Question

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

Answer 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

Answer 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

Answer 3

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

Answer 4

Actually I get it, thanks again