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, we first need to derive the velocity function from the acceleration function. Given \( a(t) = 0.8t \), we integrate to find the velocity: \[ v(t) = \int a(t) \, dt = \int 0.8t \, dt = 0.4t^2 + C \] Using the initial condition \( v(0) = 0 \), we find \( C = 0 \). Thus, the velocity function is: \[ v(t) = 0.4t^2 \] Next, we integrate the velocity function to find the position function: \[ s(t) = \int v(t) \, dt = \int 0.4t^2 \, dt = \frac{0.4}{3}t^3 + D = \frac{0.4}{3}t^3 + D \] Using the initial condition \( s(0) = 4 \): \[ s(0) = \frac{0.4}{3}(0)^3 + D = 4 \implies D = 4 \] Thus, the position function is: \[ s(t) = \frac{0.4}{3}t^3 + 4 \]