Answer
$\frac{2}{5}$ m/s.
$-\frac{4}{125}m/s^2$.
Work Step by Step
Step 1. Given the equation of motion $s(t)=\sqrt {1+4t}=(1+4t)^{1/2}$, we have: velocity $v(t)=s'(t)=\frac{1}{2}(1+4t)^{-1/2}(4)=2(1+4t)^{-1/2}$, and acceleration $a(t)=v'(t)=2(-\frac{1}{2})(1+4t)^{-3/2}(4)=-4(1+4t)^{-3/2}$
Step 2. At $t=6$ sec, we have $v(6)=2(1+4(6))^{-1/2}=\frac{2}{\sqrt {25}}=\frac{2}{5}$ m/s.
$a(6)=-4(1+4(6))^{-3/2}=-\frac{4}{25\sqrt {25}}=-\frac{4}{125}m/s^2$.
