# Homework Help: Figuring Velocity from Acceleration

1. Jul 21, 2008

### Matt_B

Ok, this is something we have been working on, and two of us agree and one disagrees on this method in calculating the final velocity with a known distance traveled. Here is what we have come up with.

The mass is .0106 lbs the force is 6lbs, distance of acceleration is .090" starting from a complete stop.

now using F=MA i come up with 566 feet per second per second. First off, is this right?

Ok taking my 566 the length of acceleration is .090" so i take my 566 multiply by 12 to get my inches per second per second.

So now I am at 6792 inches per second per second. My acceleration distance is .090"

I multiply 6792 by .090 and reach my velocity of 611.28 inches per second since i stop pushing at .090" and now the mass is released and traveling on its own. To figure back to feet per second i just divide 611.28 by 12 and the result is 50.94 FPS.

Did i figure all of this correct?

Matt

2. Jul 21, 2008

### Matt_B

BTW guys, I am not even in school. So this really has nothing to do with homework in anyway. This is something we are trying to figure out at work :)

3. Jul 21, 2008

### Staff: Mentor

The 566 ft/s^2 looks okay for the acceleration, but to find the velocity, you need to know the time that the force is acting on the object. V = Vo + at.

Also, I think in another thread you mentioned a ball or pinball or something. A rolling ball has angular momentum that you have to take into account. Some of the energy from the push goes into the rolling energy, so the translational velocity is less than if the ball is not rolling.

4. Jul 21, 2008

### Matt_B

Well see that is my whole problem, i do not have the time. I know the distance that the force is pushing it, and i know the force being exerted, also know the mass of the object. In this particular scenario, i have no way of knowing time. Is there anyway to figure out the time from that information?

Assuming my acceleration is linear, which i believe it is.
i start at 0ft/s^2 and after one second i have accelerated to 566. Well i know that my acceleration distance is .090" and after that the acting force is taken away. Here is my main issue that i am stuck on. if i must convert that to time to find velocity, how do i accomplish that? My general thinking would be that when i reach .090" i would be traveling at 50.94 ft/s considering that i am no longer accelerating. And i completely missing something here? I do realize that this theory would be in a perfect vacuum free of any sort or friction or drag. But is my basis correct?

That was not exactly the best comparison, think more of the acceleration of the plunger without the ball even there. Only that say the centerpiece of the plunger is released free of the spring when the spring is stopped short. Does that make more sense?

5. Jul 21, 2008

### LowlyPion

I would suggest that you take the distance and convert to the time.

$$d = (1/2)at^2$$

Then you can figure velocity from

V = at

6. Jul 21, 2008

### Matt_B

Ok, so then i am at $$.09 = (1/2)*566*t^2$$
which would then turn into $$.09= 283t^2$$
so result would be $$.000318=t^2$$ is that correct?
do i then take the square root of the .000318?

so i am getting time at .017833 then to get velocity i $$V=566*.017833$$ ending in 10.09 ft/s ?

and i am assuming that unit of time is seconds, correct?

7. Jul 21, 2008

### LowlyPion

I thought you said it was .09".

8. Jul 21, 2008

### Matt_B

is this correct now?

9. Jul 21, 2008

### LowlyPion

I scratched it out as 2.9 ft/sec. Looks close enough.

10. Jul 21, 2008

### Matt_B

yeah i got 2.95 ft/s thank you alot guys, you are a HUGE help! We really appreciate it!

11. Jul 21, 2008

### alphysicist

Hi Matt_B,

I think this acceleration is not right. It is not true that a 1-pound force acting on a 1 pound mass gives an acceleration of 1 ft/s^2. With your force in pounds and your mass in pounds, Newton's law needs a conversion factor, so your equation above would be:

$$F = \frac{W}{g} a$$
where W is the mass in pounds, and g is the gravitational acceleration, which has a value of just over 32 ft/s^2.

In other words, one pound of force acting on a (roughly) 32 pound mass gives that mass an acceleration of 1 ft/s^2.

(Just like in SI units we might say that a 1 Newton force acting on a 9.8 Newton mass gives the mass an acceleration of 1 m/s^2; with the understanding that a "9.8 Newton mass" means the mass that gives a weight of 9.8 N near the earth's surface.)

12. Jul 21, 2008

### Matt_B

So what you are saying is that with my given information it would look something like this:

$$6=\frac{.0106}{32.174}a$$

which breaks down to $$6= .000329a$$
would would end up being 18,237 $$ft/s^2$$ ?

that in no way seems correct, i mean we are talking that after roughly 1 second the thing would be traveling 6x faster than a rifle bullet? LOL, idk, maybe i miss understood

13. Jul 21, 2008

### alphysicist

That looks like the right acceleration to me. If you like, you can check it by converting to SI units of newtons, kilograms, and m/s^2 and then convert the final result to ft/s^2.

About your question of speed, it might be difficult for the force to be able to act for a full second--you say the force only acts over a distance of a small fraction of an inch.

(This is a really large force to be acting on such a small mass--it would be similar to a 90000 pound force acting on a 150 pound person.)

14. Jul 22, 2008

### Matt_B

Correct me if i am wrong, but recalling gravitational acceleration, i thought that mass did not come into play? It goes back to the jr high theory that in a perfect vacuum, if you drop and bowling ball and a feather at the same height at the same time, they will hit the ground at the same exact time.

I am having a hard time wrapping my head around how a 6lb spring can get a 4.8 gram object moving at 18,000 ft/s in one second when a rifle putting out over 60,000 psi accelerating a 150 grain bullet over a distance of 22" can only get it up to 3,000 ft/s

I will do the math on that figure though over my distance. As I said, i am a tad bit lost here, so anything yall are telling me helps out greatly :)

15. Jul 22, 2008

### Matt_B

Ok, so over my distance i am getting 16.54 ft/s

ok, now i am really lost! LOL

16. Jul 22, 2008

### alphysicist

That sounds right; I'm not sure how that applies here though. Can you explain more about how you are relating that to this problem?

We are just talking about the acceleration when the force is 6lb. If we assume the force is constant and acts over a distance of 0.09inches=0.0075 feet (it is 0.09 inches, right?), the force would only be acting for a time of a little under a millisecond, and the object would only be moving somewhere around 16.5 feet/second.

Also, if the thing providing the force is a spring, the force will not be constant. The force acting will vary from 6lbs to zero lbs over the distance. (I am imagining a horizontal spring, no friction, etc.)

17. Jul 22, 2008

### alphysicist

Sorry, I did not see that you had posted already before I submitted. However, that's the speed I got in my calculation in my last post.

18. Jul 22, 2008

### Matt_B

Ok, since earths gravity is the constant here... if you change location to where gravity is different, lets say the moon... is it safe to say that the acceleration of an object will also change when accelerating perpindicular to gravitational force?

19. Jul 22, 2008

### WiFO215

The moon does not accelerate perpendicular to the Gravitational force. The gravitational force is inwards. The Gravitation IS the acceleration.

20. Jul 22, 2008

### LowlyPion

Right you are. Sorry I missed that. I just focused on the snag over distance/time without going all the way back. There is that 32 factor to address. Good catch alphysicist. That's why they pay you the big bucks.

21. Jul 22, 2008

### LowlyPion

No. This is why you are working with the mass. Once you determine the mass in 1 gravitational field, then you know it's effect in a different gravity - its weight - where F = m * g. But forces perpendicular to the direction of gravity will act the same since you are using mass and not its local weight.

22. Jul 22, 2008

### Matt_B

you misunderstood my question... if i am accelerating an object, and the line of acceleration is perdendicular to gravity.

Lets say that the force of gravity is pulling down in the vertical direction. and i am using my spring to accelerate in the horizontal axis.

23. Jul 22, 2008

### Matt_B

Ok thanks, that makes complete sense :)

I am currently working on how to phrase my next major question... keep a look out for it! I think this one is a doosey

24. Jul 22, 2008

### Matt_B

Actually let me ask this first

My understanding is that if you take one object and drop it from a given height, then you make that object 2x as long and drop it from the same height, that even though the mass is 2x as much, the force at the moment of impact is identical, but the change is that the duration of impact last 2x as long. Is this correct?

25. Jul 22, 2008

### D H

Staff Emeritus
No. You misunderstood this key post:
Newton's second law of motion is commonly expressed in the form F=ma. That is not what Newton said. He said force is proportional to the product of mass and acceleration. In other words, F=kma. Newton's second law, F=kma (or F=ma with a good choice of units) is universal. It applies on the surface of the Earth, on the surface of the Moon, all points in between, and all points further out. Different representation sytems (e.g., English and metric) will have different proportionality constants, but the proportionality constant for any one system of units is the same everywhere.

The constant of proportionality k in F=kma can be made to have a numerical value of one by a good choice of units for force, mass, and acceleration. This is exactly the case with the metric system, where one newton is defined as the force needed to make a one kilogram object accelerate at one meter/second2. This is not the case with the English system if one represents force in pounds-force, mass in pounds-mass, and acceleration in feet/second2. If you want to use these units you have to use the more general expression F=kma, where k=1/32.1740486 pounds-force/pounds-mass*seconds2/feet. An alternative is to express mass in slugs, where 32.1740486 pounds (mass) is one slug. With mass in slugs, one can once again use F=ma. The mass of that 0.0106 pound (mass) object represented in slugs is 0.00032946 slugs.

An easy way to solve this problem is to use the work-energy relation. Applying a 6 pound (force) over a distance of 0.090 inches yields 0.045 ft·lbf of work. By the work-energy relation, this means the object's kinetic energy changes by 0.045 ft·lbf:

$$\frac 1 2 m \Delta v^2 = 0.045\,\text{ft}\cdot\text{lbf}$$

or

\begin{aligned} \Delta v &= \sqrt{2\, \frac{0.045\,\text{ft}\cdot\text{lbf}}{0.00032946\,\text{slugs}}\\ &= \sqrt{273.176\,\text{ft}^2\text{sec}^2} \\ &= 16.53\,\text{ft}/\text{sec} \end{aligned}