How Do Two Accelerating Soccer Players Collide?

[kholdstare121]

I'm taking high school physics right now, and we're doing kinematics in one dimension.
Well I've been stuck on this word problem for days and stumbled across this
forum on a google search.
Thought maybe you could help. Anyway it is:
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Two soccer players start from rest, 48m apart. They run directly toward each other, both players accelerating. The first player has an acceleration whose magnitude is 0.50 m/s^2. The second player's acceleration has a magnitude of 0.30 m/s^2.
(a) How much time passes before they collide?
(b)At the instant they collide, how far has the first player run?
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I've tried to tackle this as any other algebra distance=rate*time problem (but with some physics equation involved)
I attempted to get the final velocity of both runners by using the equation
Vf^2=vi^2+at
where vf=final velocity and vi=intitial.
I used '48m' as the value for distance when I did this[/color]
Then when I get my final velocity I'd figure out the average velocity for both runners.
Then tackle the problem by adding their individual rate*time together to get rt(of the first guy)+rt(second guy)=48m ,Attempting to get the time they collide from that.
BUT, apparently by plugging in 48m in the physics equation gave me a bad final velocity, therefore a bad time.[/color]
I know the correct answer for part a is 11 seconds, but don't know how it's obtained. Please I would be greatful if anyone could help.
I'll also try to clear this up if anyone's confused.
Thanks.

 

[Israfil]

solution

Hey :)

Both players are accelerating, so the equation for both their movement is
first player s_1 = 1/2*a_1*t^2
2nd player s_2 = 1/2*a_2*t^2
with a_1 = 0.5 m/s^2 and a_2 = 0.3 m/s^2
you know that s_1 + s_2 = 48m

Now you sum up both players' equations and receive:
s_1 + s_2 = 1/2*t^2*(a_1+a_2) where t is the only onknown, so you get t!

Then just insert the value of t in the first player's equation and you'll get s_1 which answers the 2nd question.

have fun :)

 

[kholdstare121]

Oh wow!
I feel REALLY dumb right now.
Was the solution that simple?
I just worked it out and got the right answer(so it is that simple )
I was going through every other equation, missing the easy connection with this one.
Thanks a lot

 

[pandabento09]

Hey :)

Both players are accelerating, so the equation for both their movement is
first player s_1 = 1/2*a_1*t^2
2nd player s_2 = 1/2*a_2*t^2
with a_1 = 0.5 m/s^2 and a_2 = 0.3 m/s^2
you know that s_1 + s_2 = 48m

Now you sum up both players' equations and receive:
s_1 + s_2 = 1/2*t^2*(a_1+a_2) where t is the only onknown, so you get t!

Then just insert the value of t in the first player's equation and you'll get s_1 which answers the 2nd question.

have fun :)

I'm completely lost. so it would be s1+s2= 1/2t^2 (0.5 +0.30) ?
48 = 1/2 t^2 (0.5 +0.39)
what did i do wrong/?

 

[Redbelly98]

That looks correct (other than it's 0.30, not 0.39, for one of the accelerations).

If I were setting this up, I would have started by writing

0 = 48 - 1/2 t^2 (0.5 +0.30),

which results in the same thing.