Acceleration and force

1. Nov 19, 2009

RohansK

We all must have come across the specification of ' Zero to Maximimum Speed' ie called Acceleration Time or 'Pick Up ' of the bike. I would call it the 'Response Time' for the Maximum Speed. It is the ('0 to 120/100 or so km/hr in 6 secs' ) or something similar.
This means that the bike would take a time period of 6 secs to start from 0 km/hr ie standstill to attain the maximum possible speed of 120 km/hr.
I could not understand what is the reason for this time lag. wait....wait , you guys may think that I dont even know that friction on ground, the air drag etc would play a role, but then actualy my question is " Why does it take time for a accelerating a mass ( here the bike) to a said speed starting it from standstill"

1. "Why cant be the process instantenous?
2. Why is this response time needed?

3. I actually wanted to very minutely understand the concept of Acceleration which we all have taken for granted to be dv/dt ie rate of change of velocity. But my question is " How do we decide the response time -- here the term dt during the change of velocity of a mass from V1 to V2 , if the process becomes almost instantenous. Wait i will give an example of a instantenous velocity change

Example :
Consider a block of mass M with wheels moving on rails. The block is connected to another smaller tray behind it with a welded rod between the two ie the block and the tray. The tray is also on wheels on the same rail, such that the block and the tray form one unit and the tray moves along the block when the block moves as they are joined by the rod.

On the tray is a system which ejects a jet of air which hits the rear face of the block applying a Force to it which causes the block to move forward ( and along it moves the tray, so that the same intensity and flowrate of the jet is maintained as the distance between the block and the tray remain the same throughout).

So as the same Force is exerted on the block throughut during one particular flowrate of the jet, then just imagine that if the jet is first locked by plate at the nozzle mouth and all of a sudden the plate is removed, then the jet would hit the block all of a sudden with all the Force INSTANTENOUSLY, ie dt = 0, causing the block to gain a speed of say V1 m/s, then , " How would we calculate dv/dt in such a case., when there is no 'Response Time' ie the the full intensity of the force is applied in one shot. First say the jet exerts F1 and in another case F2 such that F1> F2 and both act in a sudden ejection ie one -go zzoommm- fashion on the block with full intensity in one shot. In Second case velocity is sayV2.

My question is :

1. Does it always require a Response Time to gain a certain velocity from zero, and if so WHY? ( even if you say Inertia, then why some time, why not instantenous? ) Neglect friction, drag etc. Recall the bike example.

2. How to calculate the Acceleration in such a case if the rise in velocity is almost Instanenous, then does the rate of change in velocity from zero to respective max. speeds for F1 and F2 be measurable? ie how quickly has the velocity changed from zero to max -- is it measurable? If Yes or No then How? because a = dv/dt

Can anyone explain this in detail Please what happens to the VELOCITY and ACCELERATION on a micro level of TIME GAP when the Force ( F =M.A) acts on the body INSTANTENOUSLY with NO LAG at all. And note that the FOrce ( of te Jet) use do cause this motion has an ACCELRATION term in it ie F=M.A

Sorry for the very long question, but couldnt help as had to give a detailed example.

Please respond to it ansd clear a very Findamental Concept for me.

-Rohan.

2. Nov 19, 2009

gmax137

If the force changes instantaneously from zero to some value, then the acceleration will also change (instantaneously) from zero to some value (=F/m); and the velocity will also begin to change. If the force is constant at the non-zero value, the acceleration will be constant and the velocity will continue to increase as long as the force continues to be applied. In your bike example, the maximum velocity is attained when the restraining forces (friction, drag etc) are equal to the motive force. If you choose to 'neglect' these restraining forces there is no maximum velocity, ie, it will simply continue to increase.

As to your train/tray example, either I misunderstand what it looks like, or you do not understand Newton's laws: in particular, the squirting 'jet' pushes back on the 'tray' as much as the jet impinging on the block pushes the block forward. In other words, the device does not move down the tracks if the nozzle is attached to the tray.

3. Nov 19, 2009

chrisphd

If you have a body we will name A, and you apply a force to the body, of F, then that means body A will change velocity according to the formula F = mdv/dt. So if F = 10 Newtons, and mass of body A = 1kg, then body A will accelerate at 10m/s/s.

4. Nov 19, 2009

RohansK

@gmax : Good you pointed out my mistake regarding the train/tray example. So we will correct it.
And for the first example ( ire the bike one), I would say that ONLY consider it from Inertia point of view for a while, just forget the effect of friction and drag,(dont just bring them in consideration - their absecnce or presence both, see only Mass and Inertia and FOrce )

CORRECTION : Let the tray be seperated from the block, but make a provision such that the tray also moves with the same velocity as that of the block/train and the Jet system remaining the same. Is it OK now.

5. Nov 19, 2009

chrisphd

When you apply a Force to a body, you cause the body to accelerate, at a rate given by acceleration = Force/mass of body. Therefore the only way to have a sudden change in velocity (ie, dv/dt being infinite) is to have acceleration infinite, and thus have an infinite force applied to the body.

In your train example, the applied force is not infinite so the block will not change velocities immediately.

6. Nov 19, 2009

xxChrisxx

^ this ^

and for emphasis
F = ma

To get to a speed instantly you require infinite acceleration, which requires infinite force.

1. You can't have infinite force.
2. Even if you could, when you accelerated instantly you'd be crushed, and squiched by said force. (ie it would kill you.)

Put it this way, when someone falls from a high building and hits the ground he expereinces a max deceleration on the order of 100's of G. We know this kills people instantly and is very messy. Infinite G accceleration... well, its a whole lot worse.

Last edited: Nov 19, 2009
7. Nov 19, 2009

Staff: Mentor

Infinitely worse, in fact.

8. Nov 19, 2009

gmax137

Look at it this way:
If you apply a force of 1/32 pounds to an object with a weight of one pound, it will accelerate at one foot per second per second, and it will continue that acceleration as long as you continue to apply the force. In one second it will have a velocity of 1 foot per second. At 1/2 second it had a velocity of 1/2 foot per second. At 0.0001 seconds it had a velocity of 0.0001 foot per second. And at 0.000 0001 seconds it had a velocity of 0.000 0001 foot per second. The change in velocity IS instantaneous - it is just that the velocity continues to increase as long as you continue to apply the force.

9. Nov 19, 2009

FoxCommander

Lets look at this in terms of limits. You want the time between when you are at rest to what ever speed you want to be instantaneous, deltaT=0. we will use this equation: Velocityfinal= Velocityinitial+Acceleration(Time,T). If your Vf(final velocity) is your desired velocity and Vi is your starting velocity, in this case 0. So that means that you could rewrite this as Vf/T=A(acceleration). Lets take the limit of T as it approaches-->0 The answer you would get is that the A is INFINITY and so the force needed is F=M(some mass)*A. and any number times INFINITY is INFINITY. So you will always no matter how much force you apply to your bike have a delay time.
Sincerely FC

10. Nov 20, 2009

RohansK

@ gmax and @ Fox Commander: Both of you, Thanks for replying.

What both of you have explained is exactly what was haunting me and been in my mind ( and thats why I qouted the above examples tosee if anyone could come up with the idea of an Impact Force) BUT the actual question comes after this.... That what would when be in case of an IMPACT where the Force is known to be applied for a very very small interval ( may be of the order of very few microseconds) like a hammer hitting a nail, or a huge pendulum striking a small mass. And the mass literally flying off Instantenously .... no Delay Time whatsoever ( may be a very few microseconds )

Now what would be the acceleration and the whole concept of Force nearing Infinity due to acceleration approaching infinity as dt tends to zero (in an IMPACT time interval --> 0).

Expecting further assistance from you both and everryone else on the case of Acceleration and rising of speed during an Perfect Impact Collision.

@ xxChrisxx :

Could you render your expertise here. You can use an even bigger font if you wish to.

@ jtbell:

Anything better than this would be welcomed by me, so how bout replying in terms of Physics rather than making this forum a Joke. And with this approach you are a Mentor ??? :uhh:

Can you do something better so we can stick to pure Physics.

Guys expecting a reply from the serious people here.

11. Nov 20, 2009

xxChrisxx

Look its very simple, the harder you hit it the faster it accelerates accoring to newtons second law. That IS the physics that dictates how and why this happens. To get something moving and accelerating, you have to apply a certain impulse. (force*time). As time reduces, force must increase.

It's just like counting up, to get from 1 to 5, you have to go through 2,3,4. No matter how fast you do this, you still have to do it to be counting up.

This is a problem that arises from discretising a continuous operation, you can consider the change from 1 to 2 instant if you are counting in seconds. But when accelrating it has to accelrate through all the speed inbetween 1 and 2. The smaller the timestep dt, the more accurate you have to be with your timing system to register the smooth acceleration. The main point is however small you make the timestep, the timestep never becomes zero. So you can never have an instantanous change of velocity.

For purposes of practicality, when you hit something with a hammer, it may appear to instantly fly off. But it still has to accelrate up to its speed. But if you measures it more carefully, you'd see it acclerate and change velocity over time.

Last edited: Nov 20, 2009
12. Nov 20, 2009

sganesh88

13. Nov 20, 2009

FoxCommander

Couldnt have put it better my self. One more thing, you can never reach infinity, thats why it is a constant. therefore as time is approaching zero its getting infinitely smaller but never reaches 0. I dont know if that helps but just thought you should know
Sincerely,
FC

14. Nov 20, 2009

gmax137

Well, if you're interested in impact forces and resulting motion that's kind of a new subject, at least in my mind. Your original post started off with 0 to 60 mph times for bikes - that is certainly not an impact problem.

Impact forces have been discussed here in countless threads, apparently they are hard to understand. Typically the confusion arises from failure to distinguish between force and impact (force*time). Or inability to accept that determining the time duration is a difficult problem. If I understand your recent posts, you are asking about the time part, and how small or short it can be. In the real world of extended objects, everything takes some finite time to happen. Say you hit a golf ball with the club - it looks like the ball gets hit and just takes off, right? But if you saw a slo-motion movie you'd probably see the club contact the ball, the ball would deform, and at some point during that blubbery deformation, the ball would begin to move off the tee.

In the idealized world of point particles there is no spatial extension to accommodate such deformation, and you might talk yourself into instantaneous events and infinite accelerations, but you need to realize that these may be figments of the oversimplified model you are working with.