Distance Covered by Puck A at Collision

[Kalie]

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 . Simultaneously, each puck is given a quick push and they begin to slide directly toward each other. Puck A moves with a speed of = 2.70 , and puck B moves with a speed of = 5.10 .

What is the distance covered by puck A by the time the two pucks collide?

Really I just need to know how to set it up becuase each time I do this problem I get a negative answer when finding time and well time can't be negative can it?

Please help me...my brain hurts

 

[quasar987]

What have you tried? Show your work first so we can comment on it.

 

[Kalie]

Well what I have been doing is that I know that X1a=X1b and the equation of x1a= x0a + V0a (t1) and x1b= x0b - V0b (t1)
X1a= 2.70 t1
X1b+ 18-(-5.10) t1

2.70 t1= 18+5.10t1
2.70t1-5.10t1=18
-2.4t=18
t=-7.5
so then puck a traveled a -20.25m

so really that what I have been doing I tweeked it to make it positive but puck a can't travel farther than 18 m so really I got stuck in a hole after that...It has been a couple of days

Have I been approaching it wrong?
I setted up a list of knowns and unknowns

 

[quasar987]

The approach is perfect. You are using the equations of kinematics under a null force with puck A at the origin and puck B at x=18.0. However, the general equation of kinematics (i.e. the "model" applicable for any more under a null force) is

$$x(t)=x_0+v_0t$$

But you have been using the (wrong) equation $$x(t)=x_0-v_0t$$ for particle B. This is where your error comes from.

 

[Kalie]

Thank you so much! Its correct now! Just goes to show that the mastering physics dude was wrong
Thank you

 

[andrevdh]

Your initial equation for puck b seems fine to me, since it's x-coordinate will decrease with time. For puck a the x-coordinate will be

$$x_a = 2.70t$$

and for puck b

$$x_b = 18 - 5.10t$$

their x-coordinates will be the same at a time given by

$$2.70t = 18 - 5.10t$$

 

[quasar987]

Well to avoid confusion, (and also because that is what the maths behind the derivation of the kinematics equation says), we take

$$x(t)=x_0+v_0t$$

as the base equation, and let $v_0$ itself be either positive or negative wheter the motion is in the positive x-direction or negative x-direction respectively.

 

[verty]

This question seems far easier to do as a ratio, because the distances traveled is in the same ratio as the speeds, so you can use (V_a/V_t) * S_t.