Two cars, A & B, are travelling along the outside lane of a motorway at a speed of 30.0 m/s.

Question

Car B has reaction time 0.9s, brakes with the same constant deceleration as A,-6.15 m/s^2

Instead of only slowing down to 22.0 m/s. the cars had to stop. Lines are added to the graph of v-t to final velocity 0 for both

Explain why a collision is now more likely?

(initial separation is 7.2m)

Answer 1

I can provide a none technical answer...

Car B is behind car A, both traveling at the same speed, but A brakes 1st. Car B's reaction time is 0.9 seconds, (as a demonstration, lets say 1 second).  So after one second, car A has braked and is now traveling at a slower speed - lets say 25m/s, but car B is still traveling at 30m/s

So the longer they have to brake for, the more chance B has of running into the back of A because he will always be traveling that bit faster.

1 second: A = 25m/s B= 30m/s

2 seconds: A = 20m/s B = 25m/s

3 seconds: A = 15m/s B = 20m/s

and so on.

These are just examples, and not the correct data, but:

Because they brake at a constant rate, and B will be traveling faster than A, sooner or later, B will run into the back of A