Question #2566c

1 Answer
Jun 14, 2016

Answer:

No. Even if he was sober he would manage to drop to a speed of #16.5m/s# before hitting the child.

Explanation:

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.