Question

the position of a particle moving along a coordinate line is s = √(8 + 4t), with s in meters and t in s. the velocity at t = 2 sec is 1/2 m/sec. (simplify your answer. type an integer or a fraction.) the acceleration at t = 2 sec is m/sec². (simplify your answer. type an integer or a fraction.)

Explanation:

Step1: Find the velocity function

The velocity $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=\sqrt{8 + 4t}=(8 + 4t)^{\frac{1}{2}}$. Using the chain - rule $\frac{d}{dt}(u^n)=nu^{n - 1}\frac{du}{dt}$, where $u = 8+4t$ and $n=\frac{1}{2}$. So $v(t)=\frac{1}{2}(8 + 4t)^{-\frac{1}{2}}\times4=\frac{2}{\sqrt{8 + 4t}}$.

Step2: Find the acceleration function

The acceleration $a(t)$ is the derivative of the velocity function $v(t)$. $v(t)=2(8 + 4t)^{-\frac{1}{2}}$. Using the chain - rule again, $a(t)=2\times(-\frac{1}{2})(8 + 4t)^{-\frac{3}{2}}\times4=-\frac{4}{(8 + 4t)^{\frac{3}{2}}}$.

Step3: Evaluate acceleration at $t = 2$

Substitute $t = 2$ into the acceleration function. When $t=2$, $a(2)=-\frac{4}{(8+4\times2)^{\frac{3}{2}}}=-\frac{4}{(8 + 8)^{\frac{3}{2}}}=-\frac{4}{(16)^{\frac{3}{2}}}=-\frac{4}{64}=-\frac{1}{16}$.

Answer:

$-\frac{1}{16}$