The velocity ($\mathrm{mi}$/$hr$) of a drone flying in the air is given by $v\left(t\right)=12+4t^2$ for $0\le t\le 4$ hours. Let $s\left(0\right)=0$.
Determine $s\left(t\right)$ for $0\le t\le 4$.

• A. $s\left(t\right)=\frac{4t^2}{3}+12t$
• B. $s\left(t\right)=8t$
• C. $s\left(t\right)=\frac{4t^3}{3}+12t$
• D. $s\left(t\right)=\frac{8t^3}{3}+4t^{}$

Answer: (C)

Verified step by step guidance

1. Step 1: Recognize that the problem involves finding the position function s(t) from the given velocity function v(t). The position function is the integral of the velocity function with respect to time, as velocity is the derivative of position.
2. Step 2: Write the given velocity function as v(t) = 12 + 4t^2. To find s(t), compute the integral of v(t) with respect to t: ∫v(t) dt = ∫(12 + 4t^2) dt.
3. Step 3: Break the integral into two parts for simplicity: ∫(12 + 4t^2) dt = ∫12 dt + ∫4t^2 dt. Now, compute each term separately.
4. Step 4: For the first term, ∫12 dt, the result is 12t (since the integral of a constant is the constant multiplied by the variable of integration). For the second term, ∫4t^2 dt, use the power rule for integration: ∫t^n dt = (t^(n+1))/(n+1). Applying this, ∫4t^2 dt = (4/3)t^3.
5. Step 5: Combine the results of the two integrals to get the position function: s(t) = (4/3)t^3 + 12t + C, where C is the constant of integration. Use the initial condition s(0) = 0 to solve for C. Substituting t = 0 into s(t), we find C = 0. Thus, the final position function is s(t) = (4/3)t^3 + 12t.