Question

The position equation for a particle is s(t)=√t^3+1 where "s" is measured in feet and "t" is measured in seconds. Find the acceleration of the particle at 2 seconds

A) 1ft/sec^2
B) 2/3 ft/sec^2
C) - 1/108 ft/sec^2
D) None of these

Answer 1

if you mean
s = (t^3 + 1)^0.5
then
s' = v = 0.5 (t^3+1)^-0.5 ( 3 t^2) = 1.5 t^2 (t^3+1)^-0.5
s" = a = 1.5t^2(-0.5)(t^3+1)^-1.5 (3t^2) + 3t(t^3+1)^-0.5
at t = 2
a = -3 (9)^-(3/2) + 6(9)^-(1/2)
= -3 /3^3 + 6/3
= -1/9 + 2 = 17/9
check my arithmetic !!!!

Answer 2

For s" I get
(3t(t^3+4)) / (4(t^3+1)^(3/2))
so, s"(2) = 2/3