Brain Teaser

[F6 RAPID]

  Honest Gaza said:

  TurboDewd said:

F6 Rapid and 4runner are correct.

AT the moment the engines are switched off no more force is causing either vehicle to move.  Newton's first law of motion applies:

I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.

Each car had a force being applied to it (the engine) and that force stopped, they would then uniformly slow to a complete stop since the effects of air resistance and friction slow the cars down.

I'm not saying I am corect...but I disagree with your answer.

Force = Mass x Acceleration.....and therefore, the Forces would be different due to the diferences in Acceleration. (This was argued in an earlier thread when discussing the merits of having better brakes for those cars that have the ability to accelerate quickly. While some said a car doing 100kmh will need the same braking power as any other car doing the same speed, and weighing the same, others argued that whichever car accelerated fastest would require additional brake power appeared valid)

Someone who knows much more than me can tell me if I am full of it.

You are trying to make a problem W-A-Y more complicated than what it is.

Simply, at the end of the accelerating and what not (and all of that has absolutely no bearing on the actual question) , two cars (identical, same tyres, same road, no incline - ALL THINGS BEING EQUAL) are both doing an IDENTICAL 100km/hr. The question is, which one will stop first?

They should theoretically stop at exactly the same time. They are both coasting to a stop from 100km/hr with all external factors the same. How they accelerated there is absolutely irrelevant.

[Honest Gaza]

  F6 RAPID said:

  Honest Gaza said:

  TurboDewd said:

F6 Rapid and 4runner are correct.

They should theoretically stop at exactly the same time. They are both coasting to a stop from 100km/hr with all external factors the same. How they accelerated there is absolutely irrelevant.

But that is my point....."How they accelerated there is ABSOLUTELY TOTALLY RELEVANT"

[F6 RAPID]

  Honest Gaza said:

  F6 RAPID said:

  Honest Gaza said:

  TurboDewd said:

F6 Rapid and 4runner are correct.

They should theoretically stop at exactly the same time. They are both coasting to a stop from 100km/hr with all external factors the same. How they accelerated there is absolutely irrelevant.

But that is my point....."How they accelerated there is ABSOLUTELY TOTALLY RELEVANT"

I (obviously) agree!

[seventytwo]

The second car is turned off the moment it hits 100kph, as it is accelerating hard to this point the car would begin to deccelerate (accelerate slower) at that point but its velocity would continue to take it past 100 until the force of gravity and air and rolling resistance absorb the energy imparted to the the car, the car stops accelerating completly and velocity drops till the car comes to a stop.

First Car max speed 100kph

second Car 100kph + X kph

All things being = it takes longer to slow from a higher speed.

"My brain Hurts" Ralph Wiggim

[seventytwo]

  Blonk said:

True - The second car is travelling under acceleration so if it is switched off at the exact moment it hits 100 it will travel above 100 until the forces of air resistance and friction casue it to slow...

I.E.

Car 1

The car travelling at 100 (lets say its a Expensive Daewoo cause 100 is stalling speed for a ford) =

F = ma

F = ?

m = 1600kg

a = 0 as car is travelling at constant 100kmh (vu = 100 , v = 100)

Therefore without taking into consideration the resistance caused by friction and drag(as these will be equal on both vehicles at the 100kmh mark and negate each other) The positive force applied on car 1 when shut off would be F=1600x0

F=0

The coefficients of drag and friction would slow the vehicle from 100

Car 2

Accelerating from standstill to 100 (because it is a Expensive Daewoo we will say 8 second) therefore

F = ?

m = 1600

a = v-u/t

Therefore

v=100

u= 0

t = 8

100-0/8 = 12.5

a= 12.5 m/s squared

1600 x 12.5 m/s squared

Therefore

Vehicle 2 hitting 100 still has 20000N or 20 kn still being applied as it hits 100.. The curve would be parabolic but as the force decreased back to zero the speed would creep above 100.

Contrary to this is the laws of momentum that state

Momentum = p

p= mv

m= 1600

v = 21 m/s

p = 33600 kg m/s

Therefore both vehicles have the same momentum at 100 kmh therefore stopping at the same distance from shutdown.

The formula for Conservation of Momentum

Ptot =P'tot  m1V1 + m2V2 + m3V3 = m1V'1 + m2V'2 + m3V'3

<{POST_SNAPBACK}>

72........ What He said

[Dagabond]

I have nothing to add other than wanting to know what happened to M in the other Brain Teaser

[phantomchic]

the way the question is written and knowing that car 2 is on a increasing acceleration pattern and car 1 is maintaining a speed.

car 2 is not travelling a genuine 100 k's but it has merely reached 100k's perhour at the time the engine is switched off.

therefore more force is being applied to car 2 to reach the 100ksper hour, so the momentum of vehicle 2 is greater than that of vehicle 1. this in turn affects the stopping time and distance.

the force momentum of each vehicle is different at the time of engine kill because of the obvious speed achieved timing differences

car 1 is cruising steady at 100km/k

Car 2 accelerates at full throttle

I think car 1 would stop first

physics without all the letters and maths signs?

Shazzy

(now also suffering brain pain.. )

[crazed1]

First car stops first, simle really!

[Honest Gaza]

  TurboDewd said:

To some of us the answer will be obvious, but Im surprised by the number of ppl Ive asked this who get it wrong.

Still surprised ??????

[F6 RAPID]

Damn, I AM!!