# Kinematic Equations

Hey guys how's it going. My physics teacher gave us a challenge problem for extra credit, me needed the extra credit i decided to come here for help.

## Homework Statement

here is the link to the problem itself. http://www.modelingphysics.org/consta/challenge.pdf

I have actually done number 1, i can't cant seem to figure out how exactly to piece together 2. I have to come up with a algebraic equation to show exactly when the carts will collide.

Given is:
Cart A: Starts at a position of 0, velocity of zero, and an unknown a acceleration
Cart B: Starts at unknown position, velocity of zero, and unknown acceleration.

## Homework Equations

Exact problem is in link. As far as equations go it uses the kinematic equations

## The Attempt at a Solution

My physics teachers told us to put the variables in a list then plug them into a kinematic equation to solve for a unknown value then.
Cart A:
X=0
V1=0
A=?
Vf=?
t=?
Cart B:
X=?
v1=0
Vf=?
t=?
A=?
_______________________________________________________________________________
A= Acceleration
T= Time
V1= initial velocity
X= Position
VF= Final velocity

Im only confused as this involves puting them into a equation that has them colliding and that is where im stuck.

I would gladly appreciate the help. Thanks

When the two carts collide, they are at the same position at the same time. What are the kinematic equations involving time and position?

I used x=v1+1/2at^2.
As the equation. How would I go from there?

gneill
Mentor
The carts have different initial positions (only cart A starts at position 0). How would you modify your equation to account for this? Write the equation for each block.

i set the both equations equal to each other, i also removed v1 from the equation as in both equations it is 0.

so x=1/2at^2=x=1/2at^2.

i also know that Cart B has to go negative for them to meet. But how would i account if for instance, cart b did not move, or vice versa? i also have to answer what would happen if both of them did not move. and my equation is supposed to answer all of these questions. thanks for the help so far.

gneill
Mentor
The problem is, you haven't accounted for the fact that cart B does not start at x=0. The equation for cart B requires a term for its initial position. Yes, the initial velocities are both zero, so those terms vanish.

If either cart doesn't move then its acceleration will be zero. No problem! The math will take care of it. It just means that the cart that does move will eventually run into a cart that's standing still.

Last edited:
Ok i got Xb= Xf-Xb for the cart B position term.

gneill
Mentor
Ok i got Xb= Xf-Xb for the cart B position term.

Can you explain how you got that? Isn't Xb one of the unspecified givens (parameters) for the problem? And surely Xf is the final position that you're trying to locate.

You need to work the initial position of cart B, which is a parameter of the problem, into the equation for the position of cart B. In other words, the version of the formula you've used to find the cart positions over time is not the most general version of that formula. The more general version has terms for both initial velocity AND initial position, so things don't all have to begin at the origin.

Can you explain how you got that? Isn't Xb one of the unspecified givens (parameters) for the problem? And surely Xf is the final position that you're trying to locate.

You need to work the initial position of cart B, which is a parameter of the problem, into the equation for the position of cart B. In other words, the version of the formula you've used to find the cart positions over time is not the most general version of that formula. The more general version has terms for both initial velocity AND initial position, so things don't all have to begin at the origin.

Alright, how would i go about finding this formula? Sorry i'm not really keen on physics lol.

gneill
Mentor
Alright, how would i go about finding this formula? Sorry i'm not really keen on physics lol.

Somewhere in your notes or textbook you'll have a list of kinematic equations. The one you've quoted already, (1/2)at2, is a trimmed-down version of the one you want -- you want the version that includes the initial position term.

Gold Member
There are 4 kinematic equations

ok would i use: vf^2= vi^2+2ax

that's the only one i found with position in it. if so, how would i apply this?

gneill
Mentor
ok would i use: vf^2= vi^2+2ax

that's the only one i found with position in it. if so, how would i apply this?

If your course text and notes are failing you, you could always try the Physics Forums' Introductory Physics Formulary. Look under the Basic Equations of 1-D Kinematics for Displacement & Time.

• 1 person
If your course text and notes are failing you, you could always try the Physics Forums' Introductory Physics Formulary. Look under the Basic Equations of 1-D Kinematics for Displacement & Time.

Oh ok thanks a lot for that.

so i would use: x=x0+v0t+(1/2)at2?

What exactly does the 0 mean next to the x and v ? now that i have this how would i go about apply this?

gneill
Mentor
Oh ok thanks a lot for that.

so i would use: x=x0+v0t+(1/2)at2?
Yes. Note its similarity to the equation you first proposed. The difference is in the extra term representing the initial positional offset, ##x_0##.
What exactly does the 0 mean next to the x and v ? now that i have this how would i go about apply this?
The "0" subscripts designate that they represent initial conditions. Sometimes an "i" for "initial" is used instead. Think of it as representing conditions that exist at time t = 0.

As to how you apply it, that's what solving the problem is all about. What values would you put to each of the terms for the two carts involved?

so for cart B: xf=xb+v0t+(1/2)aBt2
and cart A: xf=0+v0t+(1/2)aAt2

Do i need to set them equal to each other now?

gneill
Mentor
so for cart B: xf=xb+v0t+(1/2)aBt2
and cart A: xf=0+v0t+(1/2)aAt2

Do i need to set them equal to each other now?

Both carts are starting from rest. What does that tell you about the ##v_0## parameters for each?

Make sure that you adjust your equations to comply with the chosen coordinate system. You want cart B to be moving towards the left (towards the origin), so its acceleration term should be negative. Once that's settled, yes, equate them and carry on.

P.S. You can create subscripts and superscripts using the x2 and x2 icons on the Advanced editing panel if you want to dress up your equations.

xf=xb+v0t+(1/2)-aBt2=xf=0+v0t+(1/2)aAt2

so like this?

gneill
Mentor
xf=xb+v0t+(1/2)-aBt2=xf=0+v0t+(1/2)aAt2

so like this?
Sort of. Can you clean that up so that there's an equation (only one "=" sign), and do something about the minus sign in the middle of the cart B acceleration term; either place the "-aB" in parentheses or move the minus to the outside of the term. As it stands it looks like the minus sign is between two separate terms, which of course would be incorrect.

xf=xb+v0t+(1/2)-(aB)t^2-0-v0t-(1/2)aAt2

like this? still not sure how exactly to combine them together to have 1 = sign.

gneill
Mentor
xf=xb+v0t+(1/2)-(aB)t^2-0-v0t-(1/2)aAt2

like this? still not sure how exactly to combine them together to have 1 = sign.

Nooooo. You have two equations, both of which are of the form xf = <something>.

Now, you want to find out when both equations produce the same value for xf. So you equate the <something>s. With the resulting equation you determine the time that the equation is satisfied (solve for t).

So I have to combine both equations to equal time? I'm just confused on how to set them up.

gneill
Mentor
Both equations are functions of time. For what value(s) of t are they equal?

Would it be xf-something=xf-something?

gneill
Mentor
Would it be xf-something=xf-something?

The xf's should not appear at all.

If xf = <something1> and also xf = <something2>, then you write <something1> = <something2>.

Let's be a bit more specific. You have one equation for the position of cart A. Say:

xA(t) = f(t)

And you have a separate equation for the position of cart B. Say:

xB(t) = g(t)

You want to know when (for what value of time t) xA(t) = xB(t). So you equate the functions:

f(t) = g(t)

and solve the resulting expression for t.

Once you have a value (or expression) for t, you can plug that into either of the original functions to yield xf for that time.

So would I get something+f(t)=something+g(t)?

Or would it just be something =something or xf=something + g(t)?

gneill
Mentor
So would I get something+f(t)=something+g(t)?

??? The f(t) and g(t) comprise a different example to illustrate what needs to be done. The f(t) and g(t) ARE the somethings.

Oh ok my bad. So the final equation will just be something=something?

gneill
Mentor
Oh ok my bad. So the final equation will just be something=something?

That's the idea.

The "something"s are the expressions for the cart positions with respect to time.

Alright so I got
Xb+v0T+(1/2)-(aB)t^2=0+v0T+(1/2)aAT^2

gneill
Mentor
Alright so I got
Xb+v0T+(1/2)-(aB)t^2=0+v0T+(1/2)aAT^2

Be careful with your formatting. That minus sign in the middle of the acceleration term on the left side looks like an operator dividing two terms. Move the minus sign outside of the term. Use the same case for the same variables; it looks like t and T are separate variables.

What are the initial velocities of each cart?

The initial velocity is both 0. Ok I replaced the + with a - on the left and removed parenthesis around ab

gneill
Mentor
The initial velocity is both 0. Ok I replaced the + with a - on the left and removed parenthesis around ab

So if the initial velocities are both zero, what values will you plug in for the ##V_0##'s in your equation?

Remove them since are 0?