Question

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

Explanation:

Step1: Recall velocity - position relationship

Velocity $v(t)$ is the derivative of position $s(t)$. Given $s(t)=\sqrt{8 + 4t}=(8 + 4t)^{\frac{1}{2}}$. Using the chain - rule $\frac{d}{dt}[u^n]=nu^{n - 1}\cdot u'$, where $u = 8+4t$ and $n=\frac{1}{2}$.
$v(t)=\frac{d}{dt}(8 + 4t)^{\frac{1}{2}}=\frac{1}{2}(8 + 4t)^{-\frac{1}{2}}\cdot\frac{d}{dt}(8 + 4t)$

Step2: Differentiate the inner function

$\frac{d}{dt}(8 + 4t)=4$. So, $v(t)=\frac{1}{2}(8 + 4t)^{-\frac{1}{2}}\cdot4=\frac{2}{\sqrt{8 + 4t}}$

Step3: Evaluate velocity at $t = 2$

Substitute $t = 2$ into $v(t)$. When $t = 2$, $8+4t=8 + 4\times2=16$. Then $v(2)=\frac{2}{\sqrt{16}}=\frac{2}{4}=\frac{1}{2}$

Answer:

$\frac{1}{2}$