A body on a spring moves along a line with the acceleration a = -9x (SI units), where x is its…

Step 1. Integrate. $\int_0^v v\,dv = \int_2^x -9x\,dx$ gives $\dfrac{v^2}{2} = -\dfrac{9}{2}(x^2 - 4)$, so $v^2 = 9(4 - x^2)$.

Step 2. At x = 0. $v^2 = 36\ \text{m}^2/\text{s}^2$, so $v = 6\ \text{m/s}$.

Step 3. At x = 1 m. $v^2 = 27\ \text{m}^2/\text{s}^2$, so $v = 5.20\ \text{m/s}$.