# Uncertainty physics problem

## Homework Statement

Drivers that come to a stop leave different amount of gaps between their car and the car in front. It was found that the average gap was 1.45m, but as the values varied, the uncertainty was 25cm. It was also reported that the car is 5.1 ± 0.5m in average. What is the range of distances from the bumper of the first car to the back bumper of the last one, if the cars were to line up when it comes to a stop?

N/A

## The Attempt at a Solution

Add the avg gap and the avg length of the car?

You need the range of distances. So this would be the shortest possible distance to the longest.

Firstly, how many cars are there?

Average length of a car is 5.1 ± 0.5m
Average stopping gap is 1.45 ± 0.25m

So you take the smallest possible values of those two and add up to give the minimum distance and you take the largest possible values to get the maximum.

Thank you very much. Can you give me an example if 10 cars were to line up?

Well for ten cars it would be ten times the shortest possible length of a car (5.1 - 0.5) and nine times the smallest stopping gap (1.45 - 0.25).

Now, I'm assuming the first car doesn't have a stopping gap to another vehicle here. Otherwise it would be ten times the stopping gap.

thanks!