Question

Question #13b33

Answer 1

I assume you don't take reaction time (normally about 1 second) into account. Otherwise you have to add `10m//s` times reaction time to the total.

If you decelerate at `3.6m//s^2` then your stop-time is:

`(10m//s)/(3.6m//s^2)~~2.78sec`

Your average speed during this time is `((10-0)m//s)/2=5m//s`

So you travel `2.78s*5m//~~13.89m`

Answer 2

Meneer has already given a good answer so here's one using the equation of motion:

You travel 13.9m

`v^2=u^2+2as`

`v=0m//s`

`u=10m//s`

`a=-3.6m//s//s`

`2as=0-(10xx10)=-100`#

`s=(-100)/(2xx(-3.6))=13.89m`