#### Answer

$$v(6)=\frac{2}{5}(m/sec)$$
$$a(6)=-\frac{4}{125}(m/sec^2)$$

#### Work Step by Step

$$s=\sqrt{1+4t}=(1+4t)^{1/2}$$
a) Find the formula for the particle's velocity and acceleration:
- Velocity: $$v=s'=\frac{1}{2}(1+4t)^{-1/2}(1+4t)'=\frac{1}{2\sqrt{1+4t}}\times4=\frac{2}{\sqrt{1+4t}}$$
- Acceleration: $$a=v'=\frac{(2)'\sqrt{1+4t}-2(\sqrt{1+4t})'}{1+4t}=\frac{0\sqrt{1+4t}-2s'}{1+4t}$$ $$a=\frac{-2\times\frac{2}{\sqrt{1+4t}}}{1+4t}=-\frac{4}{(1+4t)\sqrt{1+4t}}$$
b) For $t=6$ sec: $$v(6)=\frac{2}{\sqrt{1+4\times6}}=\frac{2}{\sqrt{25}}=\frac{2}{5}(m/sec)$$ $$a(6)=-\frac{4}{(1+4\times6)\sqrt{1+4\times6}}=-\frac{4}{25\sqrt{25}}=-\frac{4}{125}(m/sec^2)$$