# Motion Word Problems Help!

1. Feb 10, 2006

### forevergone

I seem to have some irregularities with a specific problem which goes like this:

"A stone is thrown upward from the edge of a cliff, with a speed of 70m/s. If, when it is 120m above the hand on its return journey, a second stone is dropped over the edge of the cliff, when and where will the stones be together?"

A nice little illustration:
http://img381.imageshack.us/img381/8154/testing6dv.th.jpg [Broken]

A graph to illustrate the velocities of the 2 stones:
http://img403.imageshack.us/img403/3606/testing6cg.th.jpg [Broken]

Now first, what I did was calculate the final velocity of stone #1:

2a + Δd = Vf^2 - Vi^2
2(10m/s^2) + (Δd) = Vf^2 - (-70m/s)^2
Vf^2 = 2500
Vf = 50m/s

Using this final velocity, I found the distance of stone #2:

2a + Δd = Vf^2 - Vi^2
2(10m/s^2) + (Δd) = (50m/s)^2 - (0m/s)^2
Δd = 2480m

Now, the only time in which the stones can be together is when their final velocities are the same. Knowing this, we use this equation:

Δd = (vi)(Δt)+1/2a(Δt)^2
(2480m + 120m) = (50m/s)(Δt) + 1/2(10m/s)(Δt)^2 (In this step, the total distance is the distance from stone 2 but also added from stone 1)
52/5 = Δt^3
Δt = 2.2

The time in which the two stones meet is 2.2s but the answer is 2.4.

Did I do something wrong in my process, or is my approach completely wrong?

Last edited by a moderator: May 2, 2017
2. Feb 10, 2006

### Staff: Mentor

>the only time in which the stones can be together is when their final velocities are the same.

I don't think that's true. Doesn't the first stone pass the 2nd stone? BTW, since the acceleration is constant, shouldn't the graphs of the velocities versus time have a different shape other than straight lines? The stone coming down from above has a downward velocity when the 2nd stone is dropped with zero velocity. Since the acceleration is constant, what can you say about the relationship between the two stone velocities?

3. Feb 10, 2006

### Staff: Mentor

This equation is incorrect. (If you check units, you'll see it can't make any sense.) The equation you are thinking of is:
$$v_f^2 = v_i^2 + 2 a \Delta y$$

As berkeman pointed out, this is not correct. The stones are together when they are at the same position. You have to find when and where that happens.

Hint: Find expressions for the position of each stone as a function of time.

4. Feb 10, 2006

### forevergone

@Berkeman - If the two accelerations are constant, does that mean the velocities are the same?

And yes, that's the equation I meant. Sorry for the mix-up.

When you say express the position as a function of time, do I equate the two equations afterward to determine the time in which the position of the stones are the same?

Last edited: Feb 10, 2006
5. Feb 10, 2006

### Staff: Mentor

Nope. Only if their initial velocities were the same. The acceleration is the change in velocity with respect to time, right? The derivitave of the position with respect to time is the velocity, and the derivitave of the velocity with respect to time is the acceleration. If the initial velocities are different, then they will stay different. The one with the higher initial velocity will keep getting faster and faster compared to the other one, which is why the first falling stone eventually catches and passes the dropped stone in this problem.

Yes. Write y1(t) and y2(t), and set them equal to each other to find the time t when the two y positions are equal. You'll most likely end up with a quadratic algebraic expression to solve....

6. Feb 10, 2006

### forevergone

Ok, so does this look right?

I foudn the final velocity first because it's shared among stone #1 and stone #2.

2aΔd = Vf^2-Vi^2
2(-10m/s)(120m) = Vf^2 - (70m/2)^2
Vf^2 = 2500
Vf = 50m/s

Using this, we set up the first equation to express the position as a function of time for stone 2:

Δd = ViΔt + 1/2aΔt^2
x = (0)Δt + 1/2(-10m/s^2)Δt^2
x = (5m/s^2)Δt^2

But also for stone 1:

Δd = ViΔt + 1/2aΔt^2
(-x+120) = (-50m/s)Δt + 1/2(-10m/s^2)Δt^2
-x = -50m/sΔt + (-5Δt^2)+120
x = 50Δt + 5Δt^2 - 120

I just assigned the displacement value with a variable so it's easier to followthrough.

So equating the x's:

50Δt + 5Δt^2 - 120 = 5Δt^2
50Δt = 120
Δt = 2.4

So is this what you mean by expressing the positions as a function of time?

Note: I was lost midway throughout the problem in stone 1 before when determining the equation for stone 3. I forgot that it was still 120m into the air .

7. Feb 10, 2006

### lightgrav

result is okay.
you're playing a bit loose with your +'ve and -'ve signs ...
be careful in future problems to use them consistently ;
it's recommended to draw on your diagram an arrow in the "positive" direction.

8. Feb 12, 2006

### forevergone

Yup,, my sign convention. I'll definitely keep that in mind.