A car accelerates from rest to 20 m/s in 10s, stays at 20 m/s for 20s, then decelerates to rest in 5s. Calculate the total distance traveled.

The car accelerates from $0 \text{ m/s}$ to $20 \text{ m/s}$ in $10 \text{ s}$. This is the area of a triangle on a velocity-time graph.

$$d_1 = \frac{1}{2} \times \text{base} \times \text{height}$$ $$d_1 = \frac{1}{2} \times 10 \text{ s} \times 20 \text{ m/s}$$ $$d_1 = 100 \text{ m}$$

The car travels at $20 \text{ m/s}$ for $20 \text{ s}$. This is the area of a rectangle.

$$d_2 = \text{velocity} \times \text{time}$$ $$d_2 = 20 \text{ m/s} \times 20 \text{ s}$$ $$d_2 = 400 \text{ m}$$

The car decelerates from $20 \text{ m/s}$ to $0 \text{ m/s}$ in $5 \text{ s}$. This is the area of a triangle.

$$d_3 = \frac{1}{2} \times \text{base} \times \text{height}$$ $$d_3 = \frac{1}{2} \times 5 \text{ s} \times 20 \text{ m/s}$$ $$d_3 = 50 \text{ m}$$

$$D_{\text{total}} = d_1 + d_2 + d_3$$ $$D_{\text{total}} = 100 \text{ m} + 400 \text{ m} + 50 \text{ m}$$ $$D_{\text{total}} = \boxed{\text{550 m}}$$

$$T_{\text{total}} = t_1 + t_2 + t_3$$ $$T_{\text{total}} = 10 \text{ s} + 20 \text{ s} + 5 \text{ s}$$ $$T_{\text{total}} = 35 \text{ s}$$

$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$$ $$\text{Average speed} = \frac{550 \text{ m}}{35 \text{ s}}$$ $$\text{Average speed} \approx \boxed{\text{15.71 m/s}}$$

---

: Stone Dropped from a Cliff

Given:
Initial velocity ($u$) = $0 \text{ m/s}$ (since it's dropped)
Height of the cliff ($s$) = $80 \text{ m}$
Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$ (standard value)

Since the stone is dropped, its initial velocity is $0 \text{ m/s}$. We use the equation:

$$s = ut + \frac{1}{2}gt^2$$ $$80 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2$$ $$80 \text{ m} = 4.9 \text{ m/s}^2 \cdot t^2$$

$$t^2 = \frac{80 \text{ m}}{4.9 \text{ m/s}^2}$$ $$t^2 \approx 16.3265 \text{ s}^2$$

$$t = \sqrt{16.3265 \text{ s}^2}$$ $$t \approx \boxed{\text{4.04 s}}$$