Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 79 m long.
The coefficient of kinetic friction between rubber and the pavement is about 0.70. Estimate the initial speed of that car assuming a level road.
m/s
Help pleeease!
8 Answers

Let initial speed 'u'=?.
As the car finally stops , final velocity 'v'= zero
distance s= 79 m
acceleration 'a =friction force /mass= (mu) mg/m=(mu)g= 0.70*9.8=  6.86 m/s^2
Assuming constant acceleration,
v^2  u^2 =2as gives
u^2=v^2  2as
u = sq rt [v^2  2as] =sq rt [zero  2as]=sq rt [ 26.8679]= sq rt1033.8=32.922 m/s
Assuming a level road. the initial speed of that car is 32.922 m/s or 118.52 km/h

The kinetic friction is given as = F(k)= 0.7
Also given that the car has come to rest (V=0) after a distance of 79 m.
V^2  U^2 = 2.a.s where V is final velocity, U initial velocity a is the acceleration and 's' is the distance traveled during this change of velocity.
The acceleration is derived from
Retarding force is m.(a) = F(k) . m . g
Therefore a = F(k) . g = 0.7 . 9.8 = 6.86 m/s^2
As V= 0 we have U^2 = 2 (6.86) 79 = 1083.88
U^2 = 1083.88 which means U the initial velocity = 32.92 m/s
corresponding to 118.5 Kmph.
Note: The initial velocity is at the beginning of the skid mark i.e. at the point where the car wheels started skidding.

V=the square root of (2μg*d)
μ=0.70 drag coefficient of the particular pavement
g=9.8 meters/s²
d=79 meters
V=32.922 meters/second or 73.64 miles per hour
Source(s): http://www.harristechnical.com/articles/skidmarks.... 
do you mean the speed of the vehicle BEFORE it started to brake???
I need you to supply me with the weight of the vehicle, and .7 is an unusual figure, and you missed some important bits, are you making this up from the top of your head, Huh?

112.85 Km/h

Zero, when it first started off that day.

i mnot physics expert sorry i can't help.
it may depend on road roughness.

zero!!!