Question #2566c

The distance it will take for the drunk man to stop is the reaction distance plus the brake distance:

`s_(st op)=s_(react)+s_(break)`

During the reaction time, speed is constant, so the distance is:

`s_(react)=u_0*t_(react)`

`s_(react)=20*0.25`

`s_(react)=5m`

The brake is decellerative motion, so:

`u=u_0-a*t_(break)`

`0=20-3*t_(break)`

`t_(break)=20/3sec`

The distance needed to stop is:

`s_(break)=u_0*t_(break)-1/2*a*(t_(break))^2`

`s_(break)=20*20/3-1/2*3*(20/3)^2`

`s_(break)=400/3-3/2*400/9`

`s_(break)=400/3-1/2*400/3`

`s_(break)=200/3m`

The total stop distance:

`s_(st op)=s_(react)+s_(break)`

`s_(st op)=5+200/3`

`s_(st op)=71,67m`

Child is dead. Here are some bonuses:

a) What if the man was not drunk?

Reaction distance changes since the reaction time is now 0.19 sec:

`s_(react)=u_0*t_(react)`

`s_(react)=20*0.19`

`s_(react)=3.8m`

The distance now becomes:

`s_(st op)=s_(react)+s_(break)`

`s_(st op)=3.8+200/3`

`s_(st op)=70,47m`

Child is still dead.

b) What is the velocity with which the child was hit?

If the driver was drunk, after 5 meters, that means at 20,1 meters close to the child he started decellerating. The impact distance is:

`s_(break)=u_0*t_(break)-1/2*a*(t_(break))^2`

`20,1=20*t_(break)-1/2*3*(t_(break))^2`

`3/2*(t_(break))^2-20*t_(break)+20,1=0`

Solving this quadratic gives:

`t_(break)=12,24sec`

or

`t_(break)=1,095sec`

We accept the smallest value, supposing he doesn't want to reverse and run the child over again. Finally, to find the velocity:

`u=u_0-a*t_(break)`

`u=20-3*1,095`

`u=16,72m/s`

`u=60,18(km)/h`

If you do the same with a sober driver you will find the child was hit with `59,4 (km)/h`. Bottom line is, he was running too fast.