1. **Stating the problem:** We have an object starting at point O with initial velocity $V=0$. It moves for two time intervals: $\Delta t_1 = 2s$ and $\Delta t_2 = 3s$. We need to find the distances traveled from O to A during $2s$ and from A to B during $3s$. 2. **Assumptions:** Since velocity is zero at the start and no acceleration is given, we assume constant acceleration $a$ during the motion. 3. **Distance from O to A for $2s$:** Using the equation for distance under constant acceleration starting from rest: $$d = V_0 t + \frac{1}{2} a t^2$$ Since $V_0 = 0$, this simplifies to: $$d_{OA} = \frac{1}{2} a (2)^2 = 2a$$ 4. **Distance from A to B for $3s$:** At point A, velocity is: $$V_A = V_0 + a t = 0 + a \times 2 = 2a$$ Distance from A to B is: $$d_{AB} = V_A \times 3 + \frac{1}{2} a (3)^2 = 2a \times 3 + \frac{1}{2} a \times 9 = 6a + 4.5a = 10.5a$$ 5. **Summary:** - Distance O to A: $d_{OA} = 2a$ - Distance A to B: $d_{AB} = 10.5a$ Without a specific acceleration value, distances are expressed in terms of $a$.