Question

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=0.2 t ; v(0)=0, s(0)=1$$

Short answer

Answer: The position function is $$s(t) = \frac{1}{15}t^3 + 1$$.

Step-by-step solution

Find The Velocity Function by Integrating Acceleration Function

Integrate the given acceleration function with respect to time to find the velocity function: $$v(t) = \int a(t)dt = \int (0.2t)dt$$

Calculate the Constant of Integration in the Velocity Function

Integrate the equation derived in step 1 to find the value of the constant of integration, C: $$v(t) = \frac{1}{5}t^2 + C$$ Since we know that the initial velocity $$v(0) = 0$$, we can substitute it and solve for the constant C: $$0 = \frac{1}{5}(0)^2 + C$$ This gives us that $$C = 0$$. So the velocity function becomes: $$v(t) = \frac{1}{5}t^2$$

Find The Position Function by Integrating Velocity Function

Integrate the given velocity function derived in step 2 with respect to time to find the position function: $$s(t) = \int v(t)dt = \int \left(\frac{1}{5}t^2\right)dt$$

Calculate the Constant of Integration in the Position Function

Integrate the equation derived in step 3 to find the value of the constant of integration, D: $$s(t) = \frac{1}{15}t^3 + D$$ Since we know that the initial position $$s(0) = 1$$, we can substitute it and solve for the constant D: $$1 = \frac{1}{15}(0)^3 + D$$ This gives us that $$D = 1$$. So the position function become: $$s(t) = \frac{1}{15}t^3 + 1$$ Hence, the position function of the object moving along a line is $$s(t) = \frac{1}{15}t^3 + 1$$.