Question #2566c

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Question #2566c

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Answer 1 · 2016-06-14T12:34:32

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

Explanation:

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

$s_{s t o p} = s_{r e a c t} + s_{b r e a k}$ $s_(st op)=s_(react)+s_(break)$

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

$s_{r e a c t} = u_{0} \cdot t_{r e a c t}$ $s_(react)=u_0*t_(react)$

$s_{r e a c t} = 20 \cdot 0.25$ $s_(react)=20*0.25$

$s_{r e a c t} = 5 m$ $s_(react)=5m$

The brake is decellerative motion, so:

$u = u_{0} - a \cdot t_{b r e a k}$ $u=u_0-a*t_(break)$

$0 = 20 - 3 \cdot t_{b r e a k}$ $0=20-3*t_(break)$

$t_{b r e a k} = \frac{20}{3} sec$ $t_(break)=20/3sec$

The distance needed to stop is:

$s_{b r e a k} = u_{0} \cdot t_{b r e a k} - \frac{1}{2} \cdot a \cdot {(t_{b r e a k})}_{b r e a k}^{2}$ $s_(break)=u_0*t_(break)-1/2*a*(t_(break))^2$

$s_{b r e a k} = 20 \cdot \frac{20}{3} - \frac{1}{2} \cdot 3 \cdot {(\frac{20}{3})}^{2}$ $s_(break)=20*20/3-1/2*3*(20/3)^2$

$s_{b r e a k} = \frac{400}{3} - \frac{3}{2} \cdot \frac{400}{9}$ $s_(break)=400/3-3/2*400/9$

$s_{b r e a k} = \frac{400}{3} - \frac{1}{2} \cdot \frac{400}{3}$ $s_(break)=400/3-1/2*400/3$

$s_{b r e a k} = \frac{200}{3} m$ $s_(break)=200/3m$

The total stop distance:

$s_{s t o p} = s_{r e a c t} + s_{b r e a k}$ $s_(st op)=s_(react)+s_(break)$

$s_{s t o p} = 5 + \frac{200}{3}$ $s_(st op)=5+200/3$

$s_{s t o p} = 71, 67 m$ $s_(st op)=71,67m$

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:

$s_{r e a c t} = u_{0} \cdot t_{r e a c t}$ $s_(react)=u_0*t_(react)$

$s_{r e a c t} = 20 \cdot 0.19$ $s_(react)=20*0.19$

$s_{r e a c t} = 3.8 m$ $s_(react)=3.8m$

The distance now becomes:

$s_{s t o p} = s_{r e a c t} + s_{b r e a k}$ $s_(st op)=s_(react)+s_(break)$

$s_{s t o p} = 3.8 + \frac{200}{3}$ $s_(st op)=3.8+200/3$

$s_{s t o p} = 70, 47 m$ $s_(st op)=70,47m$

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:

$s_{b r e a k} = u_{0} \cdot t_{b r e a k} - \frac{1}{2} \cdot a \cdot {(t_{b r e a k})}_{b r e a k}^{2}$ $s_(break)=u_0*t_(break)-1/2*a*(t_(break))^2$

$20, 1 = 20 \cdot t_{b r e a k} - \frac{1}{2} \cdot 3 \cdot {(t_{b r e a k})}_{b r e a k}^{2}$ $20,1=20*t_(break)-1/2*3*(t_(break))^2$

$\frac{3}{2} \cdot {(t_{b r e a k})}_{b r e a k}^{2} - 20 \cdot t_{b r e a k} + 20, 1 = 0$ $3/2*(t_(break))^2-20*t_(break)+20,1=0$

Solving this quadratic gives:

$t_{b r e a k} = 12, 24 sec$ $t_(break)=12,24sec$

or

$t_{b r e a k} = 1, 095 sec$ $t_(break)=1,095sec$

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

$u = u_{0} - a \cdot t_{b r e a k}$ $u=u_0-a*t_(break)$

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

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

$u = 60, 18 \frac{k m}{h}$ $u=60,18(km)/h$

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

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Question #2566c

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