Help How Do I solve Problems Like These- NOT HW- KINEMATICS

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Help!!! How Do I solve Problems Like These- NOT HW- KINEMATICS

KINEMATICS

Hey guys grade 11 physics student here,
just need a little help!

how do i solve questions that involve two missin variables?

For Example:

the uss enterprise suddently accelerates for 6.0s reaching a final velocity of 350m/s[e] the displacement for this interval was 9000m [e]

what was its intial velocity
what was its acceleration

I know the formulas but i can't use any because I don't have the variables

ALL THESE QUESTIONS ARE CONSTANT ACCELERATION

please don't post the answer but can you please show me how to go about solving them?

Thank You Very Much
 

Answers and Replies

  • #2
Doc Al
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Hint: Use two kinematic formulas and solve them simultaneously.
 
  • #3
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Along with what Doc Al said:

As you know, there are 5 different variables one can use within the 4 kinematic equations:
x, vi, vf, a, and t

Each kinematic has one variable not included. The only variable "required" for each kinematic is vi. So, you want to find one of the kinematics that does not include both vi AND a. In that case, you would be left with two unknown variables. Instead, find the kinematic that only has vi as the unknown, and solve for vi. Then you will have enough information to solve for a.

Hope this helps, and I hope I didn't say too much!
 
  • #4
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I think "solve them simultaneously" is misleading. You solve two equations, one at a time. In each one, there is only one unknown.
 
  • #5
Doc Al
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I think "solve them simultaneously" is misleading. You solve two equations, one at a time. In each one, there is only one unknown.
That entirely depends on what equations you use. Using the usual suspects--the standard kinematic formulas listed https://www.physicsforums.com/showpost.php?p=905663&postcount=2" for example--you would need to solve two equations simultaneously. None of them has only one unknown. (At least for this particular problem.)

Otherwise the problem would be too easy. :wink:
 
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  • #6
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Oh, do you not use [tex]\Delta[/tex]x = [tex]\frac{1}{2}[/tex](vi + vf)[tex]\Delta[/tex]t ?

Basically, since vavg = [tex]\Delta[/tex]x / [tex]\Delta[/tex]t
we know that [tex]\Delta[/tex]x = (vavg)[tex]\Delta[/tex]t

In cases where the acceleration is constant, vavg is equal to (vi + vf) / 2

Since, in the OP's example, acceleration is constant, we know that [tex]\Delta[/tex]x = [tex]\frac{1}{2}[/tex](vi + vf)[tex]\Delta[/tex]t

We can re-write that equation to solve for vi as follows: vi = ( 2[tex]\Delta[/tex]x / [tex]\Delta[/tex]t ) - vf

That's why I said there were four equations each with one variable "missing."
 
  • #7
Doc Al
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Oh, do you not use [tex]\Delta[/tex]x = [tex]\frac{1}{2}[/tex](vi + vf)[tex]\Delta[/tex]t ?
Hey, you're right. Forgot about that one. :approve:
 

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