# Maximum possible acceleration time period based on distance and take-off velocity?

1. Jul 20, 2009

### Researcher X

If you jump, throw a ball, or a catapult fires a shot, the projectile's acceleration surely has a maximum possible time period in which to take place? If I want to even know the acceleration I need to determine the length of this time period between velocity 0 and take-off velocity.

For example, if I throw a ball at 10m/s and the ball accelerates for the 0.8 meter of my arm before I let go, I know intuitively that this time period couldn't ever be as long as say, 100 seconds, so even though acceleration can vary independent of speed, there must be a maximum possible time period for the acceleration (and therefore a minimum acceleration) based of the distance through which the projectile accelerates and it's take-off velocity.

How do you solve for this, or am I mistaken in this assumption?

2. Jul 20, 2009

### Pengwuino

Re: Maximum possible acceleration time period based on distance and take-off velocity

I surely hope you mean 100 milliseconds and not 100 seconds! . You can determine the average force immediately since we know F=dp/dt and you know the initial momentum (0) and the final momentum (10m/s * mass) assuming you know the mass.

3. Jul 20, 2009

### Researcher X

Re: Maximum possible acceleration time period based on distance and take-off velocity

The point is that 100 seconds for that distance and final velocity is ridiculously impossible. I want to know what the limit of the timeframe would be with any given distance and final velocity.

I don't, because I'm just learning this stuff. I don't know what dp and dt stand for anyway. You'll have to explain it in laymen's terms. You say there's a formula for average force from distance and velocity?

4. Jul 20, 2009

### Pengwuino

Re: Maximum possible acceleration time period based on distance and take-off velocity

Yes, or well from velocity and the time, $$F = \frac{{dp}}{{dt}} = \frac{{\Delta p}}{{\Delta t}}$$ means the change in momentum divided by the change in time gives you the average force (the delta's stand for the change of). Momentum is simply calculated as Mass * Velocity. So if you know the mass and you know the final velocity, you know the final momentum. Of course, the beginning momentum is 0 since there is no velocity. You divide this change in momentum by the change in time, which is whatever amount of time you want to think of, and you have the average force acting on the ball.

If you want to think of what's the most ridiculous forces, just divide by 10 seconds or 100 seconds or a million seconds or whatever and you'll have what the average force is for that amount of time.

You can also determine the average force using just the velocity and distance the force is applied as well though. The kinetic energy of the ball, for a constant force is $$E = \frac{{mv^2 }}{2} = F*d$$! So what you'd do is calculate the energy using the velocity and then simply divide by the distance and you'll have your average force.

5. Jul 20, 2009

### Researcher X

Re: Maximum possible acceleration time period based on distance and take-off velocity

Is this the mass of the ball, or the mass of the person's arm which is producing the force? Thanks.

Also,

So, there is no maximum time frame for any given event?

That was my question in the first place, because I can't wrap my head around the idea of an arm, on the extreme end of this problem, taking several millenia to extend and reach a final velocity of 10,000,000mph! (Unless that arm was very very very long).

That's not impossible because nobody can do that, but impossible because you can't have that slow a rate of acceleration in that distance and still reach the final velocity stated?

Or can you? There must be a maximum time frame for this.

Last edited: Jul 20, 2009
6. Jul 20, 2009

### Pengwuino

Re: Maximum possible acceleration time period based on distance and take-off velocity

the mass of the ball is the mass in every calculation.

7. Jul 20, 2009

### rcgldr

Re: Maximum possible acceleration time period based on distance and take-off velocity

I think what is being asked here, is for a given distance of acceleration, and a specified velocity, what are the acceleration and time required to achieve this?

t = time
d = distance (given)
v = velocity (given)
a = acceleration

Assuming constant acceleration

d = 1/2 a t2
v = a t

t = (2 d) / v
a = v2 / (2 d)

8. Jul 20, 2009

### nooma

Re: Maximum possible acceleration time period based on distance and take-off velocity

The maximum timeframe would be the longest time it takes for the arm to be able to still impart the required amount of force on the ball. This is not 100s as the acceleration of the ball is done over the distance which the arm is moved.

This means the the maximum time is the arm moving the maximum distance that it can, imparting the smallest force needed for the acceleration of the ball.

Alternatively the shortest time is the smallest distance that the arm can move to impart the acceleration on the ball, this will be dependant on the maximum velocity (or force) you are able to impart using your arm.

This is all assuming a constant acceleration

9. Jul 20, 2009

### Researcher X

Re: Maximum possible acceleration time period based on distance and take-off velocity

Thanks Jeff, but I'm not clear on what the positioning of d for distance in brackets with a 2 before it (2 times possibly) means, like here: "t = (2 d) / v". I'm 100% un-knowledgable on the way maths language/formula work beyond secondary school.

I can see this "distance equals half acceleration" and then "time squared" but what's in between to determine the relation of the first to the second? Plus? Times?

"v = a t" is also typed like this. Acceleration and time are just sitting next to each other with no symbol in between. Does this mean "velocity equals acceleration x time" or squared?

Sorry.

10. Jul 20, 2009

### rcgldr

Re: Maximum possible acceleration time period based on distance and take-off velocity

For constant acceleration:

distance = (average velocity) x time

The initial velocity is 0, and the final velocity is 'v', the average, the average velocity is v/2,

d = (v/2) x t

t = (2 x d) / v

Times. Multiplication is assumed bewteen adjacent symbols in formulas.

d = (1/2) x a x t2

The implcation is "times":

v = a x t

11. Jul 20, 2009

### Researcher X

Re: Maximum possible acceleration time period based on distance and take-off velocity

Thanks. That clears it up, I think.

So, for 10m/s with a 0.8 meter distance, using the formula:

t = (2 x 0.8) / 10 = 0.16

Approximately a 10th of a second is required. Correct?

a = 10 squared / (2 x 0.8) = 62.5m/s2

So, acceleration cannot be any lower than 62.5 meters per second squared?

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

### rcgldr

Re: Maximum possible acceleration time period based on distance and take-off velocity

correct, about 1/6th of a second.

correct.