# Fnet Acceleration Problem

1. Dec 10, 2006

### XPX1

1. The problem statement, all variables and given/known data

In baseball, a pitcher can accelerate a 0.15 kg ball from rest to 90 mi/h in a distance of 1.6 m.
(a) What is the magnitude of the average force exerted on the ball during the pitch?
____N

2. Relevant equations
a=d/t
xf=xi+vixt+1/2(a)t^2
Fnet=ma

3. The attempt at a solution

I know that to find out this problem I need to find out the Fnet. I know that Fnet=ma but how do I find out the a!?

I know to figure out acceleration its distance/time but..

How do I figure out acceleration? This is getting really frustrating. It gets to 90 mi/hr in 1.6m.

2. Dec 10, 2006

### cristo

Staff Emeritus
Acceleration is *not* distance/time. Velocity is distance/time, whereas acceleration is rate of change of velocity (i.e. velocity/time)

3. Dec 10, 2006

### XPX1

Also, what does it mean by (a) what is the magnitude of the average force exerted on the ball during the pitch!?

4. Dec 10, 2006

### XPX1

I dont know what vi=?

is vi=90? or is that vf?

5. Dec 11, 2006

### Hootenanny

Staff Emeritus
The magnitude is simply the size of the force exerted on the ball; $|\vec{F}| = m|\vec{a}|$
Vi is the intial velocity, so yes, vi=90mi/h. It would however, be sensible to convert this speed into m/s.

6. Dec 11, 2006

### XPX1

Any help on how to plug this in? Now that I have the velocity, I need to find the time, dang these problems are confusing, how do I figure out the time!?

Last edited: Dec 11, 2006
7. Dec 12, 2006

### XPX1

any ideas?

8. Dec 12, 2006

### XPX1

I have

vi=90 mi/hr
mass=0.15
xf=1.6

I cant do anything with this information! Ive tried velocity versus time graphs, position versus time graphs, everything! It seems like this problem does not give me enough information to solve it!

9. Dec 12, 2006

### curly_ebhc

Change vi

I'm sorry but vi is not equal to 90 mi/hr.

The initial velocity is vi=0 (the key word is at rest).

vf=90 mi/hr but this is useless until you convert to m/s.

Now you know vi, vf, xf, xi=0, so plug into your equation, do some algebra to isolate acceleration and this is easily solved.

PS. later you'll learn work-energy and it'll save you a few steps, but for now the basics.

10. Dec 12, 2006

### XPX1

11. Dec 12, 2006

### Dorothy Weglend

v = at, as I am sure you know. Well, it's v = v0 + at. But your v0 = 0.

You can use this to eliminate the 't' in the distance equation.

Dorothy

12. Dec 13, 2006

### XPX1

That dosn't make any sense to me!

v0 = 0... Ok

v = 0+at

now how do I figure out acceleration and time!? Can somebody please do an equation of this so I can see how this works out, ive been trying to figure out this problem for 3 days now, and ive had no luck.

13. Dec 17, 2006

### cristo

Staff Emeritus
I don't know whether you've finished this question yet, but I can see that your last post hasn't been answered.

Now, before calculating anything, convert the final speed vf, into m/s.

Using your first equation, obtain an expression for the acceleration, noting that vi=0, so a=vf/t, which gives t=vf/a.

Now, use the second eqn, with xi=0, vi=0, and xf=1.6m. This gives 1.6=1/2(at^2), but from above, t=vf/a, and so we can substitute this into the expression to yield 1.6=(1/2)*a*vf^2/a^2.

From here, you can calculate a (on substituting in the converted vf) and thus go on to calculate the net force, using your third equation.

14. Dec 17, 2006

### curly_ebhc

Listen to cristo but add in one short cut. If you grab a physics book you will find an equation with no time. It looks something like this.