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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.
 
 

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