anonymous
  • anonymous
If a car travelling at vm/s takes X meters to stop, what would the stopping distance be if it was travelling at 3vm/s (assume stopping force is equal)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I remember doing this in class, but forgot which equation to use
lgbasallote
  • lgbasallote
hmm could it be \[V_2 ^2 = V_1 + 2as\] where s is the distance
lgbasallote
  • lgbasallote
correction.. \[V_2 ^2 = V_1^2 + 2as\]

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anonymous
  • anonymous
V is velocity? a is acceleration? s is displacement?
lgbasallote
  • lgbasallote
yup
anonymous
  • anonymous
Oh... yeah so that's the same one I have here as v^2=u^2+2as
lgbasallote
  • lgbasallote
yep same shiz
lgbasallote
  • lgbasallote
so first find the acceleration using v = 0 u = v s = X
anonymous
  • anonymous
0=v^2+2aX... what is a?
lgbasallote
  • lgbasallote
you solve for it
lgbasallote
  • lgbasallote
isolate a
anonymous
  • anonymous
a=-v^2/2X
lgbasallote
  • lgbasallote
right
lgbasallote
  • lgbasallote
now use v^2 = u^2 + 2as again this time use: v = 0 u = 3v a = -v^2/2X
lgbasallote
  • lgbasallote
then solve for s
anonymous
  • anonymous
starting with 0=9v^2+2aX -9v^2/2a=X -9v^2/2(-v^2/2X)=X 9X=X now what?
lgbasallote
  • lgbasallote
no...
lgbasallote
  • lgbasallote
it's not 2aX
lgbasallote
  • lgbasallote
it's 2as
lgbasallote
  • lgbasallote
X is just for the first situation
anonymous
  • anonymous
oh i see
lgbasallote
  • lgbasallote
\[0 = 9v^2 + 2as\] \[-9v^2 = 2as\] \[-\frac{9v^2}{2a} = s\] \[-\frac{9v^2}{2(-\frac{v^2}{2X})} = s\] \[\implies -\frac{9v^2}{-\frac{v^2}{X}} = s\] \[\implies 9X= s\]
anonymous
  • anonymous
Oh. THank You!
lgbasallote
  • lgbasallote
welcome

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