Rearranging equations.

  • Thread starter Serj
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  • #26
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I'm finnaly in physics and doing quite well but I am having some trouble.
d=Vi t + 1/2 a t^2 ,t=?
first I would divide both sides by Vi?
d/Vi=t + 1/2 a t^2
then divide both sides by 1/2 a? would it be d/Vi/a or d*a/Vi or d/Vi*a?
i'm going to just guess d/Vi*a (since m/s/s=m/s^2)
d/Vi*a=t +1/2 t^2 would t^2 still be multiplied by 1/2?
(square root of) d/Vi*a= t+1/2 t would the entire left side be square rooted or just the top?
t+1/2 t equals t i believe so t=(sq root of) d/Vi*a is the correct?

Second problem equation:
Vf^2=Vi^2 + 2 ad , d=?
first I would divide both sides by d?
Vf^2/2d=Vi^2 + 2a would it I still multiply d by 2?
here's were I get stuck, what do I do with Vf^2?
 
  • #27
VietDao29
Homework Helper
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[tex]d = v_it + \frac{1}{2}at^2[/tex] And you want to solve for t. Rearrange it as:
[tex]\frac{1}{2}at^2 + v_it - d = 0[/tex]. And it's a quadratic equation.
A quadratic is defined as:
[tex]ax^2 + bx + c = 0[/tex], a, b, c are already known, x is the unknown.
And its solution can be found by letting: [tex]\Delta = b ^ 2 - 4ac[/tex]
If [tex]\Delta \geq 0[/tex], then the equation has solution(s):
[tex]x = \frac{-b \pm \sqrt{\Delta}}{2a}[/tex]
If [tex]\Delta < 0[/tex], then the equation has no solution.
You can see http://en.wikipedia.org/wiki/Quadratic_equations for more information.
Here t is unknown.
---------------
For your second question:
[tex]v_f ^ 2 = v_i^2 + 2ad[/tex]
Isolate d by adding [itex]=v_i^2[/itex] to both sides.
[tex]v_f ^ 2 - v_i^2 = 2ad[/tex]
To isolate d, divide both sides by 2a:
[tex]\frac{v_f ^ 2 - v_i^2}{2a} = \frac{2ad}{2a}[/tex]
[tex]\frac{v_f ^ 2 - v_i^2}{2a} = d[/tex]
Viet Dao,
 
  • #28
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VietDao29 said:
[tex]d = v_it + \frac{1}{2}at^2[/tex] And you want to solve for t. Rearrange it as:
[tex]\frac{1}{2}at^2 + v_it - d = 0[/tex]. And it's a quadratic equation.
A quadratic is defined as:
[tex]ax^2 + bx + c = 0[/tex], a, b, c are already known, x is the unknown.
And its solution can be found by letting: [tex]\Delta = b ^ 2 - 4ac[/tex]
If [tex]\Delta \geq 0[/tex], then the equation has solution(s):
[tex]x = \frac{-b \pm \sqrt{\Delta}}{2a}[/tex]
If [tex]\Delta < 0[/tex], then the equation has no solution.
You can see http://en.wikipedia.org/wiki/Quadratic_equations for more information.
Here t is unknown.
---------------
For your second question:
[tex]v_f ^ 2 = v_i^2 + 2ad[/tex]
Isolate d by adding [itex]=v_i^2[/itex] to both sides.
[tex]v_f ^ 2 - v_i^2 = 2ad[/tex]
To isolate d, divide both sides by 2a:
[tex]\frac{v_f ^ 2 - v_i^2}{2a} = \frac{2ad}{2a}[/tex]
[tex]\frac{v_f ^ 2 - v_i^2}{2a} = d[/tex]
Viet Dao,


careful with the signs.
 
  • #29
VietDao29
Homework Helper
1,424
3
Whoops, stupid typo... :grumpy:
Yes, it should be a '-' sign instead of a '='...
Viet Dao,
 

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