The position equation for a particle is s(t)=√(2t+1) where s is measured in feet and t is measured in seconds. find the acceleration of the particle at 4 seconds. a)3ft/sec^2 b)1/3ft/sec^2 c)-1/27ft? sec^2 d)none of these

The acceleration will be evaluated as follows:
s(t)=(2t+1)^(1/2)
the velocity will v(t) is given by:
v(t)=s'(t)=1/2*2(2t+1)^(-1/2)
v(t)=(2t+1)^(-1/2)

the acceleration will be given by:
a(t)=v'(t)=s"(t)=-1/2*2(2t+1)^(-3/2)
a(t)=-(2t+1)^(-3/2)
thus to evaluate a(t) at t= 4 sec we have:
a(4)=-(2*4+1)^(-3/2)
simplifying this we get:
a(4)=-(9)^(-3/2)
a(4)=-1/27 ft/sec^2

C]

I’m not sure about that