If a car is going 58 km/h and will skid 20 m when its brakes lock, how far will it skid if it goes 120 km/h...

Question

when the brakes lock? (This question is typical on some driver's license exams.)

Answer 1

Although I highly doubt this would be on a driver's license exam, here is an easy way to figure it out:

You cannot simply do a ratio, because the question is dealing with the difference in kinetic energies of the car. The velocity is squared in the equation for kinetic energy, so it is not a linear relationship.

We can assume that the coefficient of friction is the same, even though it would probably change a little bit depending on the conditions of the road that you skidded on.

Crash investigators typically use an equation to figure out how fast a car was going, based on the length of the skidmarks and the coefficient of friction involved. We can use this equation to figure out how much further the car would go if it was travelling at 120km/h.

First we need to convert the speeds into the correct units.

58 km/h (1h/3600s)(1000m/1km) = 16.111 m/s
120 km/h (1h/3600s)(1000m/1km) = 33.333 m/s

here is the equation the crash scene investigators use:

V = velocity at time skid starts
S = skid length
Cf = coefficient of friction

(V)^2=(S)(Cf)

sometimes they multiply it by a constant, but it will cancel out in this problem anyway.

so we can find Cf by plugging in the knowns for the first situation:

(V)^2=(S)(Cf)
16.111^2=20(Cf)
Cf= 16.111^2/20 = 12.978216

now since we are assuming the Cf is constant, plug this for the second situation:

(V)^2=(S)(Cf)
33.333^2=(S)(12.978216)
S = 33.333^2/12.978216 = 85.61 m

So the car traveling 120 km/h will skid for 85.61 m,

This is over 4 times further (4.28 x) than the first car even though the speed was slightly over doubled!

Answer 2

First you have to compute the acceleration (or deceleration). Constant deceleration from 58 to 0 km/h for 20 meters:
vf^2 = 2as, or
(58km/h*1hr/3600 sec*1000 m/km)^2 = 2*20*a
a = 58*58/40 = 6.4 m^2

Now, from 120 km/h, which is 33.33 m/s, it will take 33.33/6.4 seconds to come to stop, and during that time, using
d = 1/2at^2,
we have d = 1/2*6.4*(33.33/6.4)^2
or d = 86 m

Answer 3

You can't answer this question without knowing 1) if the brakes locked immediately, and 2) how quickly the car travelled those 20 meters in the first instance. Alternatively, you could answer the question if you knew how much friction exists between the locked wheels and the road. Right now, there isn't enough information to answer this question.

Answer 4

First we can calculate the retardation by using brakes. using v^2=u^2-2as very final velocity is 0. as a effect, 60^2=2*a*20 3600=40a a=ninety To calculate the area interior the 2d case, one hundred twenty^2=2*ninety*s s=14400/a hundred and eighty=80m *made a small mistake earlier so I somewhat have purely replaced it

Answer 5

about 51 m

Answer 6

58/20 = 120/x and solve you get

41.38m