Deriving distance traveled

biggspc · Nov 23, 2011

A car can accelerate 0-60 mph in 8.6 seconds and 0-100 mph in 26.1 seconds. How far (in feet) would the car have to travel in order to accelerate from 0-81 mph?

zgozvrm · Nov 23, 2011

The acceleration rate from 0-60 mph is different than the acceleration rate from 0-100 mph. You could assume that the rate of acceleration is inversely proportional to the final speed (linear), but that really isn't the case with cars. So, with the given data, this problem is not solvable. The best you can do is find an estimate.

sophiecentaur · Nov 23, 2011

When I was first presented with the "Equations of Motion" I clearly hadn't been listening to 'Sir' well enough and I got very cross about just this point. I just thought "ow could you possibly tell?".
It took some time to take on board the qualification 'uniform acceleration'.

biggspc · Nov 23, 2011

zgozvrm said: ↑

The acceleration rate from 0-60 mph is different than the acceleration rate from 0-100 mph. You could assume that the rate of acceleration is inversely proportional to the final speed (linear), but that really isn't the case with cars. So, with the given data, this problem is not solvable. The best you can do is find an estimate.

What would the estimate be then?

Curramberus · Nov 23, 2011

Averaging the acceleration yields a result of 271.37m to reach the 81mph (36.21m/s).
It is a rough estimation, given that it should take more time to get that extra 19 mph from 81 to 100 mph than the 21 mph from 60 to 81 mph. Aerodynamic drag is proportional to the speed squared.

Curramberus · Nov 23, 2011

BTW:
271.37 m = 890.3 ft