       Physics and Car safety Answers to car collision questions  1. North^+ East > -   Fnet=Fn-Fg + Ff - Fapp                 ^                    ^                 0                     0 Fnet=Ff ma=Mmg(cancel out the masses) a=mg ^ vf*-vi*=2ad 0*- (-25m/s)*=2a(45m) -625 m*/s*/90m=90m(a)/90m -6.94 m/s*=a -6.94m/s*/(-9.81m/s*)=M(-9.81m/s*)/-9.81m/s* 0.71=M   The coefficent of friction between the road and the car's tires is 0.71.     2. North^+                                                Convert velocity from m/s to East > -                                                  km/h:                                                              50km (1h)      (1000m)= 50,000 Fnet=Ff                                                     h  (3600s)  (1km)         3600 ma=Mmg(cancel out the masses)                      =13.89m/s a=Mg a=(0.65)(-9.81m/s*) a=-6.38m/s* vf*-vi*=2ad 0*-(-13.89m/s)*=2(-6.38m/s*)d -192.9321m*/s*/-12.76m/s*=-12.76m/s*(d)/-12.76m/s* d=15.1m   The distance the car travels is 15.1 metres east.     3. North ^+ East > + Fnet=Fn+Fg+Ff+Fapp                 ^         ^                 0          0 Fnet=Fapp ma=Fapp   vf*-vi*=2ad 0*-(25m/s)*=2(a)(1.5m) -625m*/s*/3m=3m(a)/3m -208.3m/s*=a   ma=Fapp 80kg(-208.3m/s*)=-16,666.7 N The seatbelt provides a 1.7 x10^4 N force backwards.     4. North ^ - East > +   vix=5m/s(cos20degrees)      viy=5m/s(sin20degrees)      = 4.696 m/s                                =1.710m/s   d=1/2at* + vi(t) 1.5m-1.5m=1/2(9.81m/s*)t* + 4.698m/s(t)-1.5m 0=4.905m/s*(t*) + 4.698 m/s(t) -1.5m   ~Use the quadratic formula~ t= -4.698+- (squareroot of)  [4.698* -(4)(4.905)(-1.5)] /2(4.905) t= 0.25 s t= -1.21 s (extraneous root)   v=d/t d=vt d=(4.698m/s)(0.25 s) d=1.17 m The person flew a horizontal distance of 1 metre.        Home | The importance of seatbelts | The Use of the Airbag | Impulse | The use of Headrests in a Collision | Friction | Brakes | Airbag/Egg Lab | Friction Lab | Questions | Answers | Statistics | Glossary | Links   