Class Management |Help Chap 8 Begin Date: 3/26/2017 12:00:00 AM - Due Date: 10/2
ID: 1789463 • Letter: C
Question
Class Management |Help Chap 8 Begin Date: 3/26/2017 12:00:00 AM - Due Date: 10/29/2017 11:59:00 PM End Date: 12/15/2017 12:00:00 AM (11 %) Problem 1: A 85-kg person is riding in a car moving at 19,5 m/s (assume this is the positive direction) when the car runs into a bridge abutment. This problem will illustrate why the invention of the airbag dramatically improved the safety of automobiles 50 % Part (a) Calculale lhe honzonal component ofthe average lorce, in new lons, on the person if he is stopped by a padded dashboard that compresses an avcrage of 1.25 cm. Grade Summa Deductions Potential Fdash 1.28 106 4% 96% cosO asinacos0 789 OML 415161- Submissions Allempls remaining: 2 (4% per attempt) detailed view sinO atanacotan nh cosh0 4% tanh cotanhC END Degrees Radians BACKSPACL DEL CLLAR Submit Hint Feedback I give up! Hints: 400 deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback. 50% Part (b) Calculate the horizontal component of the average force, in newtons, on the person if he is stopped by an air bag that compresses an average of 14. cm All content .© 2017 Expert TA, LLCExplanation / Answer
Impulse is defined as change in momentum, dP = M dV = Favg dT
so Favg = M dV/dT.
So we need dS = Vavg dT and dT = dS/Vavg = 2 dS/ U = 0.025/19.5 = 1.28 x 10-3 s
So Favg = M dV/(2dS/U) = -85 x 19.5 /(2 x 0.0125/19.5) = -1.29 x 106 N
(Please consider negative sign in above result, that is very significant)
2. favg/Favg = M dV/dT//M dV/dt = (dt/dT) = (ds/dS); so that favg = Favg (ds/dS)
= (-1.29 x 106)(.0125/.145) = -11.14 x 104 N
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