1. **Stating the problem:** We are given the acceleration function $$a(t) = 0.012 + 0.008 \sin(0.5 t)$$ and need to find the velocity $$v(t) = \int_0^t a(s) \, ds$$ and the position $$y(t) = \int_0^t v(s) \, ds$$. 2. **Formula and rules:** Velocity is the integral of acceleration over time, and position is the integral of velocity over time. The integral of a sum is the sum of integrals, and constants can be factored out. 3. **Find velocity $$v(t)$$:** $$v(t) = \int_0^t \left(0.012 + 0.008 \sin(0.5 s)\right) ds = \int_0^t 0.012 \, ds + \int_0^t 0.008 \sin(0.5 s) \, ds$$ Calculate each integral: - $$\int_0^t 0.012 \, ds = 0.012 t$$ - For $$\int_0^t 0.008 \sin(0.5 s) \, ds$$, use substitution: $$\int \sin(k s) ds = -\frac{1}{k} \cos(k s) + C$$ Here, $$k=0.5$$, so $$\int_0^t 0.008 \sin(0.5 s) ds = 0.008 \times \left[-\frac{1}{0.5} \cos(0.5 s)\right]_0^t = 0.008 \times (-2) \left[\cos(0.5 t) - \cos(0)\right] = -0.016 \left[\cos(0.5 t) - 1\right]$$ Simplify: $$v(t) = 0.012 t - 0.016 \cos(0.5 t) + 0.016$$ 4. **Find position $$y(t)$$:** $$y(t) = \int_0^t v(s) ds = \int_0^t \left(0.012 s - 0.016 \cos(0.5 s) + 0.016\right) ds$$ Integrate term by term: - $$\int_0^t 0.012 s \, ds = 0.012 \times \frac{t^2}{2} = 0.006 t^2$$ - $$\int_0^t -0.016 \cos(0.5 s) ds = -0.016 \times \int_0^t \cos(0.5 s) ds$$ Use substitution: $$\int \cos(k s) ds = \frac{1}{k} \sin(k s) + C$$ So, $$-0.016 \times \left[\frac{1}{0.5} \sin(0.5 s)\right]_0^t = -0.016 \times 2 \left[\sin(0.5 t) - 0\right] = -0.032 \sin(0.5 t)$$ - $$\int_0^t 0.016 ds = 0.016 t$$ Combine all: $$y(t) = 0.006 t^2 - 0.032 \sin(0.5 t) + 0.016 t$$ **Final answers:** $$v(t) = 0.012 t - 0.016 \cos(0.5 t) + 0.016$$ $$y(t) = 0.006 t^2 - 0.032 \sin(0.5 t) + 0.016 t$$