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 decreasing?

a) 0<t<1
b) t>1
c) t>0
d) The velocity is never decreasing.

Thanks in advance.

Answer 1

s(t)=(t+3)(t-1)^3
s ' (t) = (t+3)(3)(t-1)^2 + (t-1)^3 (1)
= (t-1)^2 (3(t+3) + t-1)
= (t-1)^2 (4t + 8)
since we want t > 0 , what values do you think this will give us ?
Remember, for a function to be decreasing the first derivative must
be negative.

Answer 2

we have the velocity, s', above.
To find out when the velocity is increasing, we need its derivative to be positive
s" = 12(t^2-1)
so, as long as t > 1, s' is increasing.
That is, the graph of s(t) is concave up.