Where Do Emily and Grace's Bumper Cars Collide?

[SavageWarrior]

Homework Statement

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?

Homework Equations

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

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?

 

[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...

 

[LawrenceC]

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

 

[SavageWarrior]

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

 

[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.

 

[SavageWarrior]

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

 

[LawrenceC]

Yes.

 

[SavageWarrior]

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

 

[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.