Really quick question

[AgaBoogaBoo]

Doing math homework, and I'm new to driving so I don't know just yet. I'm still 15, not 16 for 2 months or so.

Anyway, why does doubling a car's speed take more than double the distance to stop?

 

[Ameesh]

because deceleration (and acceleration) doesn't neccesarly conform to linear boundries

 

[dighn]

dependence of friction on speed?

 

[AgaBoogaBoo]

Originally posted by: Ameesh
because deceleration (and acceleration) doesn't neccesarly conform to linear boundries

Thats what the data shows, but why? hehe 😛

 

[AgaBoogaBoo]

Originally posted by: dighn
dependence of friction on speed?

so at the higher speed it would have less friction requiring more time to stop so it can gain friction?

 

[dighn]

Originally posted by: AgaBooga

Originally posted by: dighn
dependence of friction on speed?

so at the higher speed it would have less friction requiring more time to stop so it can gain friction?

something like that, looking at the variables the only thing that could have a significant effect is the friction used to stop the car.

 

[m2kewl]

b/c

Energy = mass x velocity^2

fool!

😛

 

[eLiu]

Originally posted by: m2kewl
b/c

Energy = mass x velocity^2

fool!

😛

hahahaha you are funny hahaahaha lol

 

[m2kewl]

Originally posted by: eLiu

Originally posted by: m2kewl
b/c

Energy = mass x velocity^2

fool!

😛

hahahaha you are funny hahaahaha lol

well i try!!! 😛

 

[AgaBoogaBoo]

Originally posted by: eLiu

Originally posted by: m2kewl
b/c

Energy = mass x velocity^2

fool!

😛

hahahaha you are funny hahaahaha lol

Is that a real equation? I'm in Chemistry so I haven't had any of that physics stuff yet

 

[dighn]

yes that's the equation for kinetic energy, except it's too big by a factor of 2 😀

 

[AgaBoogaBoo]

Originally posted by: dighn
yes that's the equation for kinetic energy, except it's too big by a factor of 2 😀

what should it be?

<---- fumbduck

 

[Tetsuo]

Originally posted by: AgaBooga

Originally posted by: dighn
yes that's the equation for kinetic energy, except it's too big by a factor of 2 😀

what should it be?

<---- fumbduck

Your not fumbduck, your agabooga

 

[dighn]

Originally posted by: AgaBooga

Originally posted by: dighn
yes that's the equation for kinetic energy, except it's too big by a factor of 2 😀

what should it be?

sorry i mean kinetic energy = (1/2)*mass*speed^2

 

[sash1]

Originally posted by: AgaBooga

Originally posted by: dighn
yes that's the equation for kinetic energy, except it's too big by a factor of 2 😀

what should it be?

<---- fumbduck

it's E=(1/2)mv^2

 

[DrPizza]

Originally posted by: m2kewl
b/c

Energy = mass x velocity^2

fool!

😛

Uhhh, kinetic energy = 1/2 m v^2
fool!
😀
pwned!

edit: Darn... already pwned by others in this thread 🙂

 

[AgaBoogaBoo]

Originally posted by: Tetsuo

Originally posted by: AgaBooga

Originally posted by: dighn
yes that's the equation for kinetic energy, except it's too big by a factor of 2 😀

what should it be?

<---- fumbduck

Your not fumbduck, your agabooga

I know, but it sounds better than with the f and d switched around 😛 Seriously

 

[DrPizza]

Lets see if I can explain it using a simple example. Suppose you can decrease the speed of the car at a rate of 10 mph every second.
No, lets use feet per second to avoid the conversions... hours are a pita

Suppose you can slow down a car 10 feet per second every second.
Now, if the car is going 50 feet per second, it's going to take 5 seconds to stop.
If the car is going 100 feet per second, it's going to take 10 seconds to stop.

Doubling the speed doubles the time it takes to stop.

Now the tricky part. If it went from 50 feet per second to 0 feet per second smoothly, it averaged 25 feet per second. Over 5 seconds, that means it traveled 125 feet.

If it went from 100 feet per second to 0 feet per second smoothly, it averaged 50 feet per second for 10 seconds, and traveled 500 feet.

 

[m2kewl]

Originally posted by: DrPizza

Originally posted by: m2kewl
b/c

Energy = mass x velocity^2

fool!

😛

Uhhh, kinetic energy = 1/2 m v^2
fool!
😀
pwned!

edit: Darn... already pwned by others in this thread 🙂

notice the smiley...anyone could of figured that out with half a brain. *cough*google*cough*

😛

 

[dighn]

ahhh crap i was wrong. thought it said the time not the distance... oh well. in that case the distance would be proportional to the square of the speed assuming constant friction.