Does Slope Affect Car Stopping Distance?

[Kieranlavelle]

Homework Statement

A car is moving along a flat road at a constant 18ms^-1 it then applies the breaks with a force of 3600N

If this car is then moving down a hill at 18ms^-1 the same breaking force is applied is the cars stopping distance larger or smaller or the same?

Homework Equations

Eκ = 1/2 M V^2

The Attempt at a Solution

I said the same as the stopping distance is proportional to the Kinetic energy of the car is the thinking distance is not relevant so as the KE of the car is the same I assumed the stopping distance would be the same.

I then asked by brother who is rather good at maths but did not take physics and he was a bit indecisive and said it depends on how it was being marked. Can you guys shed any light on this?

 

[CWatters]

Ask yourself... What are the forces acting on the car and are they the same in both cases? Draw two free body diagrams.

 

[Kieranlavelle]

I am well aware of the component of the weight down the slope but for some reason still placed the same distance as my answer.

 

[CWatters]

I've no idea why you think it would be the same. The car on the hill has to loose both KE and PE via the brakes.

I'm not sure what your brother was thinking about either. The maximum friction force that can be applied depends on the coefficient of friction between tyre & road and the normal force. The normal force is lower for the car on the hill so the maximum friction force would be lower. However the problem statement says that both cars manage to apply the same friction force so this effect isn't a factor.

In any case it would also increase the stopping distance of the car on the hill. I can't think of any reason way the stopping distance would be longer for the car on the flat so your brothers comment doesn't make sense to me. Sorry.

 

[Kieranlavelle]

Sorry if I explained this poorly, my brother says the stopping distance is longer on the hill and the only thing that changes is the gradient of the slope and the question only accounts for the breaking force and not the friction of the tyres. if you want to know how basic this question is it was for an a level question.

 

[oli4]

Hi Kieranlavelle
Just forget the formulas and hop on a bike
Try to go uphill and see how easier or harder it is to stop your bike than on a flat road or a downhill road.
This should give you the intuition toward the right answer, which should in turn guide you to how to explain why this is so.
Once this is clear, well the problem as given has some room for improvement in terms of how it could be stated, meaningless values given, meaningful values not given (if you want to consider how a car given real life conditions is having a constant speed on a flat road and how this is translated in the downhill case etc.)

 

[CWatters]

Sorry if I explained this poorly, my brother says the stopping distance is longer on the hill

He is correct.

What would happen if the hill was very very step? Let's say it's approaching a vertical drop and the car weighs 360kg. The force of gravity acting down the hill would be almost f = ma = 360 * 9.8 = 3528N. The braking force acting up the hill is stated as 3600N. Subtract and you get the net force acting on the car...

3600 - 3528 = 72N

So the total force acting on the cars is:

On the flat: 3600N
On the hill: 72N

Fairly obvious which will slow down quicker and which will have the longest stopping distance.

If you want to think of it in terms of energy then as I pointed out earlier the brakes of the car on the hill have to dissipate both the PE and the KE where as the brakes on the flat only have to dissipate the KE.

As Oli4 points out your problem statement doesn't have enough info to calculate the exact stopping distance in each case.

 

[PeterO]

Homework Statement

A car is moving along a flat road at a constant 18ms^-1 it then applies the breaks with a force of 3600N

If this car is then moving down a hill at 18ms^-1 the same breaking force is applied is the cars stopping distance larger or smaller or the same?

Homework Equations

Eκ = 1/2 M V^2

The Attempt at a Solution

I said the same as the stopping distance is proportional to the Kinetic energy of the car is the thinking distance is not relevant so as the KE of the car is the same I assumed the stopping distance would be the same.

I then asked by brother who is rather good at maths but did not take physics and he was a bit indecisive and said it depends on how it was being marked. Can you guys shed any light on this?

Was the slope sufficiently rough so that before the brakes were applied the car was traveling at a constant 18 ms-1 as with the flat surface? ie rough enought to provide sufficient friction other than form the brakes?