Kinematics with 2 objects

1. Feb 12, 2012

SavageWarrior

1. The problem statement, all variables and given/known data
Emily and Grace are each in a bumper car facing each other 10m apart. Emily moves toward Grace at a constant speed of 2.5 m/s. Grace accelerates toward Emily at a rate of 0.5 m/s^2. Relative to where Emily started, where do the two bumper cars collide?

2. Relevant equations
x=V$_{}0$t+1/2at^2
x-x$_{}1$=V$_{}0$t+1/2at^2

3. The attempt at a solution
Grace
x=1/2at^2
x=(.5)(.5)t^2
x=.25t^2

Emily
x-10=2.5(t)
(.25t^2)-10=2.5(t)
.25t^2-2.5t-10=0
t=13.06, -3.06

x=.25(-3.06)^2
x=2.34

Can someone tell me if I did this correctly?

Last edited: Feb 12, 2012
2. Feb 12, 2012

LawrenceC

Look at your result. Emily has an initial speed of 2.5 m/s. Grace starts from 0 speed. Your answer implies Grace went a greater distance......

3. Feb 12, 2012

LawrenceC

I would have written the second equation as 10-x=2.5(t)

4. Feb 12, 2012

SavageWarrior

Well, if you write the second equation that way it doesn't change the answers you will get because you square them.

5. Feb 12, 2012

LawrenceC

But it is wrong the way you have it written. The ends don't justify the means. You want the sum of the distances moved by both to be 10 m. So you could write:

2.5t + .5at^2 = 10

If you add your two equations, you do not get the above.

The distance 2.34 m is the distance from Grace's original position. Question asks for point of impact from Emily's original position.

6. Feb 12, 2012

SavageWarrior

So, if 2.34 is the distance from Grace wouldn't 10-2.34 be the answer I am looking for?

7. Feb 12, 2012

LawrenceC

Yes.

8. Feb 12, 2012

SavageWarrior

Okay thanks, but is there a way to set up the problem so that the x calculated is from Emily's position?

9. Feb 12, 2012

LawrenceC

Sure, take the equation

2.5t + .5at^2 = 10

and solve for t just as before.

Then apply the time to

Xe = Ve * t = 2.5 * t, t=3.06 sec

which gives how far Emily moved from her initial position.