Given the following position function: (position is in ft , time is in sec) $$ \mathrm{s}(\mathrm{t})=2(4 \mathrm{t}+1)^{\frac{3}{2}} $$ Find the velocity at $$ \mathrm{t}=2 $$ $$ \mathrm{v}(2)=[?] \frac{\mathrm{ft}}{\mathrm{sec}} $$

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To find the velocity at $$ t = 2 $$ seconds for the given position function:
$$
s(t) = 2(4t + 1)^{\frac{3}{2}}
$$
we need to determine the first derivative of $$ s(t) $$ with respect to time $$ t $$, which represents the velocity $$ v(t) $$.

### Step 1: Differentiate the Position Function

Given:
$$
s(t) = 2(4t + 1)^{\frac{3}{2}}
$$

To find $$ v(t) = s'(t) $$, we'll use the chain rule for differentiation. The chain rule states that if you have a composite function $$ f(g(t)) $$, then:
$$
\frac{d}{dt} f(g(t)) = f'(g(t)) \cdot g'(t)
$$

Applying this to $$ s(t) $$:
$$
v(t) = \frac{d}{dt} \left[ 2(4t + 1)^{\frac{3}{2}} \right]
$$
$$
v(t) = 2 \cdot \frac{3}{2} (4t + 1)^{\frac{1}{2}} \cdot 4
$$
$$
v(t) = 2 \times \frac{3}{2} \times 4 \times (4t + 1)^{\frac{1}{2}}
$$
$$
v(t) = 3 \times 4 \times (4t + 1)^{\frac{1}{2}}
$$
$$
v(t) = 12(4t + 1)^{\frac{1}{2}}
$$

### Step 2: Evaluate the Velocity at $$ t = 2 $$ seconds

Plug $$ t = 2 $$ into the velocity function:
$$
v(2) = 12(4(2) + 1)^{\frac{1}{2}}
$$
$$
v(2) = 12(8 + 1)^{\frac{1}{2}}
$$
$$
v(2) = 12(9)^{\frac{1}{2}}
$$
$$
v(2) = 12 \times 3
$$
$$
v(2) = 36 \, \frac{\text{ft}}{\text{sec}}
$$

### **Final Answer**
$$
v(2) = 36 \, \frac{\text{ft}}{\text{sec}}
$$
The velocity at $$ t = 2 $$ seconds is $$ 36 \, \frac{\text{ft}}{\text{sec}} $$.

Given the following position function:
(position is in ft , time is in sec)

Find the velocity at

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Given the following position function: (position is in ft , time is in sec) Find the velocity at