the position of a particle moving along a coordinate line is s = √(25 + 4t), with s in meters and t in…

the position of a particle moving along a coordinate line is s = √(25 + 4t), with s in meters and t in seconds. find the rate of change of the particles position at t = 6 sec. the rate of change of the particles position at t = 6 sec is m/sec. (type an integer or a simplified fraction.)

the position of a particle moving along a coordinate line is s = √(25 + 4t), with s in meters and t in seconds. find the rate of change of the particles position at t = 6 sec. the rate of change of the particles position at t = 6 sec is m/sec. (type an integer or a simplified fraction.)

Answer

Explanation:

Step1: Differentiate the position - function

The position function is $s(t)=\sqrt{25 + 4t}=(25 + 4t)^{\frac{1}{2}}$. Using the chain - rule $\frac{d}{dt}(u^n)=nu^{n - 1}\frac{du}{dt}$, where $u = 25+4t$ and $n=\frac{1}{2}$. $\frac{ds}{dt}=\frac{1}{2}(25 + 4t)^{-\frac{1}{2}}\cdot4=\frac{2}{\sqrt{25 + 4t}}$

Step2: Evaluate the derivative at $t = 6$

Substitute $t = 6$ into $\frac{ds}{dt}$. When $t = 6$, $\frac{ds}{dt}\big|_{t = 6}=\frac{2}{\sqrt{25+4\times6}}=\frac{2}{\sqrt{25 + 24}}=\frac{2}{\sqrt{49}}=\frac{2}{7}$

Answer:

$\frac{2}{7}$