Question #2566c

Jun 14, 2016

No. Even if he was sober he would manage to drop to a speed of $16.5 \frac{m}{s}$ before hitting the child.

Explanation:

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

${s}_{s t o p} = {s}_{r e a c t} + {s}_{b r e a k}$

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

${s}_{r e a c t} = {u}_{0} \cdot {t}_{r e a c t}$

${s}_{r e a c t} = 20 \cdot 0.25$

${s}_{r e a c t} = 5 m$

The brake is decellerative motion, so:

$u = {u}_{0} - a \cdot {t}_{b r e a k}$

$0 = 20 - 3 \cdot {t}_{b r e a k}$

${t}_{b r e a k} = \frac{20}{3} \sec$

The distance needed to stop is:

${s}_{b r e a k} = {u}_{0} \cdot {t}_{b r e a k} - \frac{1}{2} \cdot a \cdot {\left({t}_{b r e a k}\right)}^{2}$

${s}_{b r e a k} = 20 \cdot \frac{20}{3} - \frac{1}{2} \cdot 3 \cdot {\left(\frac{20}{3}\right)}^{2}$

${s}_{b r e a k} = \frac{400}{3} - \frac{3}{2} \cdot \frac{400}{9}$

${s}_{b r e a k} = \frac{400}{3} - \frac{1}{2} \cdot \frac{400}{3}$

${s}_{b r e a k} = \frac{200}{3} m$

The total stop distance:

${s}_{s t o p} = {s}_{r e a c t} + {s}_{b r e a k}$

${s}_{s t o p} = 5 + \frac{200}{3}$

${s}_{s t o p} = 71 , 67 m$

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:

${s}_{r e a c t} = {u}_{0} \cdot {t}_{r e a c t}$

${s}_{r e a c t} = 20 \cdot 0.19$

${s}_{r e a c t} = 3.8 m$

The distance now becomes:

${s}_{s t o p} = {s}_{r e a c t} + {s}_{b r e a k}$

${s}_{s t o p} = 3.8 + \frac{200}{3}$

${s}_{s t o p} = 70 , 47 m$

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:

${s}_{b r e a k} = {u}_{0} \cdot {t}_{b r e a k} - \frac{1}{2} \cdot a \cdot {\left({t}_{b r e a k}\right)}^{2}$

$20 , 1 = 20 \cdot {t}_{b r e a k} - \frac{1}{2} \cdot 3 \cdot {\left({t}_{b r e a k}\right)}^{2}$

$\frac{3}{2} \cdot {\left({t}_{b r e a k}\right)}^{2} - 20 \cdot {t}_{b r e a k} + 20 , 1 = 0$

Solving this quadratic gives:

${t}_{b r e a k} = 12 , 24 \sec$

or

${t}_{b r e a k} = 1 , 095 \sec$

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

$u = {u}_{0} - a \cdot {t}_{b r e a k}$

$u = 20 - 3 \cdot 1 , 095$

$u = 16 , 72 \frac{m}{s}$

$u = 60 , 18 \frac{k m}{h}$

If you do the same with a sober driver you will find the child was hit with $59 , 4 \frac{k m}{h}$. Bottom line is, he was running too fast.