Deceleration of a vehicle problem

[Probie]

I do not understand vehicle deceleration.

The Attempt at a Solution

d = µg
6.687 = .7 * 9.81
d = 6.687 m/s

My question is...is it always figured out this way when dealing with vehicles.

 

[EricVT]

Can you elaborate a bit perhaps? I don't follow at all what you're trying to do or what you're confused about.

 

[Probie]

Okay, let's say you have to give an opinion on the following.
A car is approaching the rear of a stationary car. And you want to know if it will crash into the car or if there is enough distance to stop.
You need to know the deceleration rate of the car. You do not want it to skid to a stop, just a controlled deceleration. The coefficient for the road is .7 so what is the deceleration rate for the the car.

I think it gets figured out like this. But I am not sure. This is where I get confused.

d = µg
6.687 = .7 * 9.81
d = 6.687 m/s

 

[EricVT]

So the car itself is not mechanically decelerating (braking), you are essentially just traveling down the road at some constant speed and then want to know what the deceleration rate is if the accelerator is removed and the car slows down under the force of friction between it and the road surface?What is 'd'? Deceleration? If you are trying to model the car-road system as a solid mass (the car) moving on a stationary surface (the road), then your deceleration will be decided by the frictional force opposing your direction of motion and the mass of the car.

F = m*a

and also

F = u*mg

mg is the weight of the car, and assuming the road is flat this is the normal force exerted by the road onto the car. Frictional force is the product of the coefficient of static friction and the normal force. Assuming no other forces are acting on your car, you can say

m*a = u*mg
a = u*g

So your deceleration is given by the product of g and the coefficient of friction. So at least using that model of the car-road system your work seems correct.

To answer your question about whether this is ALWAYS the way to work the problem, though...in general yes. But you have to keep in mind that the normal force is not always just m*g. If the road is at some angle then you need to deal with force components in the normal direction and whatnot and do a little more work, but nothing major.

 

[TVP45]

Hint

Probie,
I assume you mean deceleration for d. You have the right idea that the maximum deceleration would be ug. This assumes all 4 wheels are braking and that the road is level but those seem reasonable. So, deceleration means slowing down FROM some velocity (speed) - what speed is that?

 

[Probie]

Thanks to both of you, the information is very helpful.

Here is the complete situation. I was asked to check on this for work.

A vehicle is stopped on the highway at a laneway. Another vehicle crests a hill at a speed of 80 km/h. Is there enough time for the second vehicle to stop before rearending the lead vehicle? ( not slam on the brakes, just an emergency controlled stop)

Here is what I did.. I used my own vehicle and time how long it would take at 80 km/h to reach the driveway. ( bit unorthidoxed but what can you do)The time from the crest of the hill to the driveway is 20.14 sec.

D= V/t

80km/h * .2777 = 22.216 m/s
22.216 m/s * 20.14sec = 447.43 meters
_______________________________________
a = µg ( a = deceleration rate)
using a µ = .75 (coefficient friction for road)
µ +- .3 (hill decline percentage)

-4.41 = .45 * 9.81
_______________________________________
D= V/t

22.216m/s / -4.41 m/s = 5.03 sec
_______________________________________

d = .5VT (d=distance , V=velocity, T = time)

55.73 = .5 * 22.216 * 5.03
55.73 meters stopping distance
add 1.5 sec for perception reaction time = 1.5 * 22.216 = 33.324 meters

55.73 + 33.324 = 89.05 meters
_______________________________________

So the distance required to stop the second car is 89.05 meters. So 447.43 - 89.05 = 358.38 meters or 16.13 sec safety margin between the stopped veh and the rear veh.

I am sorry I do not describe things very well. I hope you can understand what I have done here. ~ Thanks