# Force vs time graph

1. Oct 31, 2005

### Punchlinegirl

A soccer player kicks a ball of mass 0.45 kg that is initially at rest. The player's foot is in contact with the ball for 0.005 s. The force of the kick is shown in the Figure. What is the impulse imparted on the ball?

I know that I need to integrate to solve this problem, but how would I do that when I have no idea what the equation is?

2. Oct 31, 2005

### Päällikkö

I can't see the figure.

Anyways, when integrating, what are you actually doing, and how could you do it graphically (às in derivating - tangent of the curve)?

3. Oct 31, 2005

### Punchlinegirl

I don't know how I can find the derivative when I don't know the equation. On the graph, the y-axis is not very well labeled, so I don't know how to find a tangent line.

4. Oct 31, 2005

### Päällikkö

To find the derivative from a function graphically, you draw a tangent at the desired point of the function.

Integration is calculating the area.

5. Oct 31, 2005

### Punchlinegirl

I just don't understand how I can integrate the graph, when I don't know the original function, and the graph isn't labeled well enough to get a close enough approximation of the force.

6. Oct 31, 2005

### whozum

From what you're saying we can't really help you unless we see the graph

7. Oct 31, 2005

### Punchlinegirl

Ok.. hope this works

8. Oct 31, 2005

### whozum

You would want to integrate that, but I'm not really sure how you would go about doing so.. you could estimate it pretty well by some meticulous eyeballing but if you need an exact answer I guess what you'd have to do is model the curve. It looks pretty quadratic to me..

9. Nov 1, 2005

### verty

Well, supposing it is quadratic, one gets the friendly formula: f(x) = -1.2e9x^2 + 6e6x. One can then work out the impulse. I don't think that's what the teacher wanted, it was probably meant to be an approximation.

10. Nov 1, 2005

### Punchlinegirl

Is there a way to figure out the final velocity? I know that the initial velocity is 0 since it's at rest. I'm thinking maybe I need to use final momentum- initial momentum for the impulse since the graph isn't very easy to read.
Any advice would be appreciated. Thanks

11. Nov 2, 2005

### verty

Impulse = m * delta-v. The impulse is the area under the graph.

Last edited: Nov 2, 2005