1. **State the problem:** A glider starts descending from rest with a constant acceleration of $2\,m/s^2$ for the first 15 seconds, then continues with zero acceleration for the next 15 seconds. We need to find the total distance traveled in the first 30 seconds. 2. **Relevant formulas:** - Distance under constant acceleration: $$s = ut + \frac{1}{2}at^2$$ where $u$ is initial velocity, $a$ is acceleration, and $t$ is time. - Distance under constant velocity: $$s = vt$$ where $v$ is velocity and $t$ is time. 3. **Calculate distance during first 15 seconds:** - Initial velocity $u = 0$ (starts from rest) - Acceleration $a = 2\,m/s^2$ - Time $t = 15\,s$ $$s_1 = 0 \times 15 + \frac{1}{2} \times 2 \times 15^2 = \frac{1}{2} \times 2 \times 225 = 225\,m$$ 4. **Calculate velocity at 15 seconds:** - Using $v = u + at$ $$v = 0 + 2 \times 15 = 30\,m/s$$ 5. **Calculate distance during next 15 seconds with zero acceleration:** - Acceleration $a = 0$ - Velocity remains constant at $v = 30\,m/s$ - Time $t = 15\,s$ $$s_2 = v \times t = 30 \times 15 = 450\,m$$ 6. **Calculate total distance traveled in 30 seconds:** $$s_{total} = s_1 + s_2 = 225 + 450 = 675\,m$$ **Final answer:** The glider travels a total distance of $675$ meters in the first 30 seconds.