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
