University Physics with Modern Physics (14th Edition)

Let $v_1$ be the velocity after 5.0 s and let $v_2$ be the velocity after 10.0 s. $v_1 = (5.0~a)$ and $v_2 = (10.0~a)$ We can see that $v_2 = 2v_1$ During the second time interval of t = 5.0 s, the ball rolled a distance $d_2$ of 200 m. $d_2 = (\frac{v_2+v_1}{2})(t) = \frac{3v_1}{2}(t)$ $v_1 = \frac{2d_2}{3t}$ We can use this expression to find $d_1$. $d_1 = (\frac{v_1+v_0}{2})(t) = \frac{(2d_2)/(3t)}{2}~(t)$ $d_1 = \frac{d_2}{3} = \frac{200~m}{3} = 66.7~m$ The ball rolled a distance of 66.7 meters during the first 5.0 seconds.