Driving revisited

From my earlier post on Quadratic Equations of driving, we can establish another proof of why the breaking distance of a car increases to about 4 fold if the speed of the car increases by 2. This proof relies on two facts:

1) For a velocity-time graph, the area under the graph is equal to the displacement.
2) For two similiar figures, the ratio of their area is proportional to the square of the ratio of the corresponding lengths.

For instance, suppose a car travels at 50 km/h, he sees a boy and breaks. The reaction time is, say 1s and the car decelerates uniformly ( it need not be uniform. for simplicity, let us assume that it is uniform). one can now find the distance travelled by looking at the area under the speed time graph.

Now if he were to travel at twice the speed, at 100 km/h, how would the area under the graph look like?

From the graph, we see that the figures that form the areas for the triangles are actually similar! Let A1 and A2 be the breaking distances when inital speed is 100 km/h and 50 km/h respectively.

Then A1:A2=4:1

Thus breaking distance is increased four-fold instead of two-fold. However, we do note that in reality, the deceleration is not uniform in both cases. Nevertheless, the breaking distance will increse by more than two fold.