1. **State the problem:** We need to estimate the speed of the horse at 4 seconds using the gradient of the tangent to the distance-time curve at that point. 2. **Understanding the graph:** The graph plots distance (metres) on the x-axis and time (seconds) on the y-axis, which is unusual since speed is normally distance over time. Here, time is a function of distance, so the gradient of the curve is $\frac{d(\text{time})}{d(\text{distance})}$. 3. **Relate gradient to speed:** Speed is $\frac{d(\text{distance})}{d(\text{time})}$, the reciprocal of the gradient of the curve since the graph is time vs distance. 4. **Find the gradient of the tangent at 4 seconds:** From the graph, the tangent at 4 seconds (on the y-axis) corresponds to a point on the curve. Suppose the tangent line passes through points approximately $(x_1, y_1) = (30, 3)$ and $(x_2, y_2) = (50, 5)$ (distance, time). 5. Calculate the gradient of the tangent line: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{50 - 30} = \frac{2}{20} = 0.1$$ 6. Since the graph is time vs distance, the gradient $m = \frac{d(\text{time})}{d(\text{distance})} = 0.1$. 7. The speed is the reciprocal: $$\text{speed} = \frac{1}{m} = \frac{1}{0.1} = 10$$ 8. **Final answer:** The estimated speed of the horse at 4 seconds is $10$ metres per second (to 1 decimal place).