How do you calculate the velocity using the stopping distance?

Question

Here's there question:

A school bus full of pupils was pulling away from the curb adjacent to the school when the driver heard a squeal of brakes behind. He did not feel an impact but stopped the bus to see what had happened. At the rear of the bus, a car had come to a halt some 60cm from the bus with skid marks 64.56m in length on the dry asphalt road surface. Determine if the driver of the car was exceeding the speed limit of the urban area.
Much appreciated
Thanks

Answer 1

One method to solve this problem is to set the work done by the force of friction acting on the car equal to the change in the car’s Kinetic Energy.
The force of friction is exerting a [what we will assume to be] constant force on the car in the direction opposing motion. This force (exerting over the stopping distance) does work on the car to change [lower] its Kinetic Energy.

Force of friction = coefficient of friction * Normal Force
F_f = μ * F_n
And if the car is on a horizontal surface, F_n = weight = mg
Where g is the gravitational acceleration experience by the car (9.81 m/s^2).

Work = Force * distance
Where the Force being considered is the force of friction and the distance is the length of the skid marks.
Work = F_f * d

Kinetic Energy = ½ mv^2

Set the work done equal to the change in KE.
Since the car’s final speed is zero (at rest), the change in KE is equal to the initial KE.
Work = ΔKE
F_f * d = ½ mv^2
μ * m * g = ½ * m * v^2
We can cancel the mass term (m) out on both sides and multiply both sides by 2 to give us,
2 * μ * g = v^2
Take the square root of both sides,
v = sqrt (2μg)

But in order to solve this, we need to know the coefficient of friction between the car’s tires and the road. This will need to be given to us in the problem or we need to know where to look it up. Likely, if the question did not outright give us the value for μ, there is probably a table somewhere earlier in the chapter that lists the μ value we should use for the coefficient of friction between rubber tires and a dry asphalt road.

Answer 2

The short answer is: probably.
The long answer is: you have to do the math and assume a number since you didn't provide all the details of the problem. Let's start. This will be fun :)

Start with one of the equations of motion:

Vf^2=Vi^2+2*a*s

Where Vf=final velocity, Vi=initial velocity, a=acceleration, s=distance traveled. In your problem, Vf=0 when the car comes to a complete stop. Vi=initial velocity of the car, a=deceleration due to friction, s=stopping distance. Don't forget in your case, a is a deceleration. So it's going to be a negative number in the equation above. Letting Vf=0 and solving for Vi, you find this:

Vi=sqrt(2*a*s) (I've taken care of the minus signs and a is just a positive number now)

To get a, you bring friction into the picture:

a=F/m

where F=force of friction and m=mass of car. The force of friction can be expressed as:

F=u*m*g

where m=mass of car, g=acceleration due to gravity, and u=coefficient of kinetic friction (it's kinetic since the car left a skid mark). Put all the equations together and you get:

Vi=sqrt(2*u*g*s)

You have s=64.56m. You have g=9.8m/s^2. You have to estimate u since you didn't give me one. On dry asphalt, u=0.4 is a reasonable number. There's actually a large range, but you want to use the small number in the range to find the slowest possible speed the car was going when it hit the brakes. The larger the coefficient of friction, the faster the car was going to leave the same length skid mark. So if you want to give the driver a speeding ticket, you want to know what's the slowest the car could have possibly been going. So use a small u in the range.

Anyway, using u=0.4, you find Vi=22.5 m/s. That's about 81 Km/hr or 50 miles per hour. I don't know about you, but around where I live, school zones have 20 mph speed limit when school is in session. 50mph is way too fast for a school zone. The fact that the car stopped 60cm from the bus makes for good drama, but is inconsequential to the solution of the problem.

Answer 3

-To solve this problem we will need supose that the deceleration (a) is constant to be able to use one of the kinematic equations (V^2=Vo^2+2aΔx).

-Since we have a deceleration, 'a' will be negative; therefore changing the equation to V^2=Vo^2-2aΔx.

-If the skid marks immediately appeared when the driver of the car applied the brakes we can take the 64.56 meters as the distance traveled to come to a stop (Δx) .

-Our final velocity 'V' will be zero since at the end the car came to a stop.

-Then solve the equation for the initial velocity 'Vo' which is the one we are looking for to determine if the driver was exceeding the speed limit. We get: Vo^2=129.12a.

-Since we don't for sure the deceleration of the car 'a', we will have to do some research to look for the maximum deceleration of a car (I leave that to you). Aproximating this max 'a', we can say it's about 6 m/s^2.

-Now if we substitute 'a' for our aproximated value and solve for Vo, we get: 27.8338 m/s.

I think that maybe there is a way to come for more exact values for the deceleration using Newton's Second law (F=ma) and some algebra, but that is a bit more difficult and I don't have the time to work on it. Hope this helped.

May God Bless You.

Answer 4

Skid Mark Calculator

Answer 5

twist of destiny investigators make some assumptions in step with results of tests. They recognize a few universal issues like anticipated deceleration of a automobile (usually, some would recognize approximately bicycles) with the brakes locked. you will ought to do some finding out to characteristic information to the information you have.