Use the formula for Final velocity

[Morgan89]

If the speed of a car is increased by 75%, by what factor will its minimum braking distance be increased, assuming all else is the same



I am not sure what formulas i should use for this problem. Am i souposed to use the formula for Final velocity which is
Velocity Final^2= Velocity Initial^2 + 2*acceleration*displacement


I do not know how i would start this problem without knowledge of the variables. I am also not sure if this is even the right formula. I do not know the acceleration so i am not sure how it would work.

Help!

 

[hage567]

You already started a thread on the problem, please don't double post!

https://www.physicsforums.com/showthread.php?t=172572

So if the velocity v, of the car is increased by 75%, what is the new velocity in terms of the first velocity? That will relate the two scenarios and allow you to find the relationship between the distances. Does that make sense?

 

[Dick]

You have all of the formulas you need. Now you need to start thinking. i) What is Velocity Final? ii) True, you don't know the acceleration - but what do you know about it? Where does it come from and is it reasonable to take it to be constant?

 

[Kurdt]

I will say that you can't assume the time to stop is constant, but that is fairly easy to get in terms of initial velocity and acceleration which are constant.

 

[Morgan89]

Sorry i double posted, on my other post someone said that i had posted in the wrong section, and i was just trying to fix that. sorry.

The final velocity is zero. I understand that. But for the car to come to a stop, there would have to be a negative deceleration due to friction maybe. I do not see how it could be constant or an unnecesary figure.

 

[Dick]

Friction, clearly. What sort of friction? How would you calculate the amount of this friction?

 

[Kurdt]

I think in the case of this question you are ignoring friction and just concentrating on a constant deceleration from the cars brakes.

 

[Dick]

I think in the case of this question you are ignoring friction and just concentrating on a constant deceleration from the cars brakes.

It says 'minimum stopping distance'. I guess I was assuming that that would happen with the brakes locked. So you don't have to worry about the internals of the brake.

 

[Kurdt]

It says 'minimum stopping distance'. I guess I was assuming that that would happen with the brakes locked. So you don't have to worry about the internals of the brake.

True. But if the kinematic equations are being used then i'd assume a relatively simple braking model in which they provide a constant deceleration. If it was to do with friction then the minimum stopping distance is best worked out through considerations of energy and work.

Perhaps a bit more information on the question might be enlightening.

 

[Dick]

You could state the problem two ways.

i) Assume that the braking mechanism provides a constant stopping force independent of velocity.

ii) Leave it to student to figure that the fastest way to stop is to lock the brakes, maximizing kinetic friction. In which case you don't have to assume that it's constant. You can show that it's constant - as the poster will explain to us shortly, I hope.

You'll get the same answer either way, of course. But I prefer the second form - it has more physics in it and less arbitrary assumptions.

 

[Morgan89]

I think i figured it out! Distance and velocity are related in such a way that distance increases with the square of velocity. so if the velocity were to increase twice as much, then the distance would increase by 4 times. Therefore if my velocity increases by 3/4 the distance will increase by 9/16...Right!?

 

[Kurdt]

If you increase the velocity by 75% what do you multiply the velocity by? Its not 3/4 because you'd be reducing the velocity to 75%.