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Car safety The National Transportation Safety Bureau indicates that a person in a car crash has a reasonable chance of survival if his or her acceleration is less than 300 $\mathrm{m} / \mathrm{s}^{2}$(a) What magnitude force would cause this acceleration in such a collision? (b) What stopping distance is needed if the initial speed before the collision is 20 $\mathrm{m} / \mathrm{s}(72 \mathrm{km} / \mathrm{h} \text { or }$ 45 $\mathrm{mi} / \mathrm{h}$ )? (c) Indicate any assumptions you made.
$24 \mathrm{kN} ; 0.67 \mathrm{m}$
Physics 101 Mechanics
Chapter 2
Newtonian Mechanics
Newton's Laws of Motion
Applying Newton's Laws
Cornell University
University of Washington
Hope College
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So for the first part, the force due to the crash is given by mass of the car times its acceleration. So we have got 800 kg for the card and the acceleration is 300 m per second squared. So the force becomes 2.4 times 10 to the Five Newton way. Know that? Uh huh. If we calculate the reaction time, then the the reaction distance is given by the not times the reaction time which comes out to be 20 times 0.6. It was 12. I made dinner. Now, using the form of the the finance squared equals we initiate square minus two acceleration times displacement. The final velocity being zero. We have just X equals V initiates were over to a equals 20 square over two times 300 acres, 0.67 committed. So the total distance record will be given by the distance that you travel while before you are able to react us. Tell the X on that will come out to be who else? Foreign 67 m now, right? The assumption here is negligible here resistance and we're working with one B most
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