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Symbolic Notation question

  • #1

Homework Statement



Two diamonds begin a free fall from rest from the same height, a time "delta t" apart. How long after the first diamond begins to fall will the two diamonds be a distance "d" apart? Give your answer in terms of the variables given and "g". ( Solve for t= )

Homework Equations



Variables: "delta" t, "d", and "g"

The Attempt at a Solution


I know "g" is the gravitational constant and will have an effect on "delta" t, but I just can't figure out how to set up this equation. I know it is an form of one of the Constant-Acceleration equations, but there's like 5 of them and I don't know which one to use. Thanks in advance.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
gneill
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Hi chevymechanic, welcome to PF.

Which of the "like five" equations might be relevant? Which involves finding distance given acceleration (g) and time?
 
  • #3
I'm going to say it's: x-x0=v0t+(1/2)at2

The g constant would sub in for a and d would sub in for x-x0 correct?

Then would V0 be 0 because the objects start from rest?

Then the only variable left is "t", which is what I need to find.
 
  • #4
gneill
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I'm going to say it's: x-x0=v0t+(1/2)at2

The g constant would sub in for a and d would sub in for x-x0 correct?

Then would V0 be 0 because the objects start from rest?

Then the only variable left is "t", which is what I need to find.
You're heading in the right direction. But d is going to be the distance between the two falling objects, so you first need to use that equation (twice!) to write expressions for the individual distances for each of them. Why not call them d1 and d2.

Supposing that the first object starts falling when t = 0, then its distance equation is simple:

d1 = (1/2)*g*t2

The equation for d2 is just a bit trickier because you have to allow for the fact that it starts falling at a time [itex] \Delta t [/itex] later. How might you accommodate that in the equation?
 
  • #5
Hmm. I know that [itex]\Delta[/itex]t would be the change in time, which is final-initial. So I would have to work "t-0" into the equation somehow.

d2 = (1/2)g(t-0)2...?

I don't think that's right because "t-0" would leave me with just "t" and would be the same as the first equation...
 
  • #6
gneill
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Hmm. I know that [itex]\Delta[/itex]t would be the change in time, which is final-initial. So I would have to work "t-0" into the equation somehow.

d2 = (1/2)g(t-0)2...?

I don't think that's right because "t-0" would leave me with just "t" and would be the same as the first equation...
Ooooh, very close :wink:

[itex]\Delta t[/itex] is the offset in time for when the second object starts its fall. So in order to make the equation behave as though time [itex]\Delta t[/itex] is its zero starting point, shift the time variable by that amount:

d2 = (1/2)g(t - [itex]\Delta t[/itex])2

Note that when t = [itex]\Delta t[/itex] that the argument of the square term is zero. So the equation will behave just like the one for d1, only offset by [itex]\Delta t[/itex] in time.

Next you'll want to write an expression for d using your two equations and solve for t.
 
  • #7
If I were to write an expression for d, which would be a combination of d1 and d2, that would leave me with:

d = (1/2)g(t-[itex]\Delta[/itex])2+(1/2)gt2

Would I then set d=0 and use the quadratic formula to solve for t?

I have something similar in my notes, but this is where I usually get lost in class. :frown:
 
  • #8
gneill
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If I were to write an expression for d, which would be a combination of d1 and d2, that would leave me with:

d = (1/2)g(t-[itex]\Delta[/itex])2+(1/2)gt2

Would I then set d=0 and use the quadratic formula to solve for t?

I have something similar in my notes, but this is where I usually get lost in class. :frown:
d1 will have the larger value, since it describes the object that falls first. So you'll want to reverse the order of the terms in order to make d take on a positive value.

Since the question specifies that you want to know when the objects are a distance d apart, you don't want to set d to zero -- the result depends upon the variable d. You'll have to solve the quadratic and carry d along for the ride :smile:
 
  • #9
Well, I got the question wrong. The computer is VERY picky about what answer or format (mine was wrong) that you put in it. It was looking for:

(2d+g([itex]\Delta[/itex]t2))/(2g[itex]\Delta[/itex]t)

^^^That's supposed to be Delta t ^2nd power.

I still don't have the concept of the Constant Acceleration formulas down. My college offers free tutoring services, so I plan to take advantage of that opportunity. I never have been very well at learning over the computer. I do much better face-to-face with someone where I can ask questions as I go. Well gneill, I really appreciate the help. Thank you for sharing your knowledge and time. Have a wonderful evening!
 
  • #10
gneill
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d is the difference between d1 and d2, not the sum.
 

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