Given the following acceleration function of an object moving along a line, find the position function with the given initial velocity and position. \( \mathrm{a}(\mathrm{t})=0.8 t \mathrm{v}(0)=0, \mathrm{~s}(0)=4 \) (Type an expression using t as the variable)

To find the position function \( s(t) \) from the given acceleration function \( a(t) = 0.8t \), we will follow these steps: 1. **Find the velocity function \( v(t) \)** by integrating the acceleration function. 2. **Use the initial condition for velocity** to find the constant of integration. 3. **Find the position function \( s(t) \)** by integrating the velocity function. 4. **Use the initial condition for position** to find the constant of integration for the position function. ### Step 1: Find the Velocity Function The acceleration function is given by: \[ a(t) = 0.8t \] To find the velocity function, we integrate the acceleration function: \[ v(t) = \int a(t) \, dt = \int 0.8t \, dt \] Calculating the integral: \[ v(t) = 0.8 \cdot \frac{t^2}{2} + C_1 = 0.4t^2 + C_1 \] ### Step 2: Use the Initial Condition for Velocity We know that \( v(0) = 0 \): \[ v(0) = 0.4(0)^2 + C_1 = 0 \implies C_1 = 0 \] Thus, the velocity function simplifies to: \[ v(t) = 0.4t^2 \] ### Step 3: Find the Position Function Next, we integrate the velocity function to find the position function: \[ s(t) = \int v(t) \, dt = \int 0.4t^2 \, dt \] Calculating the integral: \[ s(t) = 0.4 \cdot \frac{t^3}{3} + C_2 = \frac{0.4}{3}t^3 + C_2 \] ### Step 4: Use the Initial Condition for Position We know that \( s(0) = 4 \): \[ s(0) = \frac{0.4}{3}(0)^3 + C_2 = 4 \implies C_2 = 4 \] Thus, the position function is: \[ s(t) = \frac{0.4}{3}t^3 + 4 \] ### Final Answer The position function is: \[ s(t) = \frac{0.4}{3}t^3 + 4 \]