Given the following acceleration function of an object moving along a line, find the position function with the given initial velocity and position. \( a(t)=0.8 t ; v(0)=0, s(0)=2 \) \( s(t)=\square \) (Type an expression using t as the variable.) v)

To find the position function \( s(t) \) given the acceleration function \( a(t) = 0.8t \), initial velocity \( v(0) = 0 \), and initial position \( s(0) = 2 \), follow these steps: ### 1. Integrate Acceleration to Find Velocity The acceleration \( a(t) \) is the derivative of velocity \( v(t) \): \[ a(t) = \frac{dv}{dt} = 0.8t \] Integrate \( a(t) \) with respect to \( t \) to find \( v(t) \): \[ v(t) = \int 0.8t \, dt = 0.8 \cdot \frac{t^2}{2} + C = 0.4t^2 + C \] Apply the initial condition \( v(0) = 0 \) to find the constant \( C \): \[ 0 = 0.4(0)^2 + C \implies C = 0 \] So, the velocity function is: \[ v(t) = 0.4t^2 \] ### 2. Integrate Velocity to Find Position The velocity \( v(t) \) is the derivative of position \( s(t) \): \[ v(t) = \frac{ds}{dt} = 0.4t^2 \] Integrate \( v(t) \) with respect to \( t \) to find \( s(t) \): \[ s(t) = \int 0.4t^2 \, dt = 0.4 \cdot \frac{t^3}{3} + D = \frac{0.4}{3}t^3 + D \] \[ s(t) = \frac{2}{15}t^3 + D \quad \text{(since } 0.4 = \frac{2}{5} \text{)} \] Apply the initial condition \( s(0) = 2 \) to find the constant \( D \): \[ 2 = \frac{2}{15}(0)^3 + D \implies D = 2 \] ### Final Position Function \[ s(t) = \frac{2}{15}t^3 + 2 \] **Answer:** \( s(t) = \dfrac{2}{15}\, t^{\,3} + 2 \)