# Calculating time

1. Sep 14, 2004

### BlackMamba

I'm am yet again confused.

I have a problem which has two soccer players starting from rest, 51m apart. They run directly toward each other. The first players acceleration is 0.50 m/s^2. The second player's accelration is 0.30m/s^2.

(a) How much time passes before they collide?

(b) At the instant they collide, how far has the first player run?

Ok so here's what I know:

Both players initial velocity is 0m/s.
Both players initial distance is 51m.
Player one has an acceleration of 0.50m/s^2.
Player two has an acceleration of 0.30m/s^2.

So I calculated the velocity of both players. Now calculating the time, is where I begin to loose focus. Do I find each players total time? If so, how does that tell me when they collide?

Thanks for any help

2. Sep 14, 2004

### Chronos

Try using this

D = Vi + (AT^2)/2 and
Vf = Vi + AT

D = distance, meters
Vi = initial velocity, meters/sec
T = time in seconds
A = acceleration, meters/sec
Vf = final velocity, m /s

Vi, as you noted is 0

So, 51 meters = A1T^2/2 + A2^T2/2
where A1 is first runners acceleration and A2 is second runners acceleration

3. Sep 14, 2004

### BlackMamba

Im not sure I understand. The first two equations you gave, I know will give me the distance and final velocity. However do I apply those first two equations to both players?

And I think I understand that the last equation will give me the time. Is that correct? However I'm having a hard time understanding what it actually reads.

Last edited: Sep 14, 2004
4. Sep 14, 2004

### Gokul43201

Staff Emeritus
How did you calculate the velocity ? Like I've asked you before, the velocity keeps changing, so which velocity are you talking about ?

Picture this scene in your head. Two players on the field... you are observing with a stopwatch...the instant they start, you start the timer....they start very slowly, but keep getting faster...the instant they collide you stop the timer. So you see; there is only one time...not a different time for each player.

The important thing to notice is that the sum of the two distances run, gives the total distnce = 51 m. Write the equations for distance of each runner in terms of time, initial velocity (=0) and acceleration. Then add these two equations and set the sum to 51. This will give you the equation for t that Chronos describes in his last paragraph.

5. Sep 14, 2004

### BlackMamba

My original thinking was to find the final velocity. But the final velocity would actually be 0, wouldn't it? As in when they collide they would actually be no longer be moving. Am I thinking this right now? So my equations for the final velocity are useless.

I did draw a picture on my paper. Trying to picture it.

Ok, I will go and work on what you have told me. I'm sure I will be back shortly though. LOL

Thank you.

6. Sep 14, 2004

### Pyrrhus

Tell me Black Mamba when you're driving a car and suddenly you hit a tree, is your final speed 0?, think about it.

7. Sep 14, 2004

### Gokul43201

Staff Emeritus
Forget about the final velocity. The conditions described (their accelerations) change after impact. We are only concerned with what happens up to instant of impact, as the given conditions apply only during this time.

Last edited: Sep 14, 2004
8. Sep 14, 2004

### Gokul43201

Staff Emeritus
If you don't knock the tree over, yes it is !

Cyclovenom, I don't mean to criticize your post, but I think it best that BlackMamba try first to understand what's happening in the given problem.