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anonymous
 3 years ago
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)
anonymous
 3 years ago
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)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I remember doing this in class, but forgot which equation to use

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm could it be \[V_2 ^2 = V_1 + 2as\] where s is the distance

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0correction.. \[V_2 ^2 = V_1^2 + 2as\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0V is velocity? a is acceleration? s is displacement?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh... yeah so that's the same one I have here as v^2=u^2+2as

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so first find the acceleration using v = 0 u = v s = X

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.00=v^2+2aX... what is a?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now use v^2 = u^2 + 2as again this time use: v = 0 u = 3v a = v^2/2X

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0starting with 0=9v^2+2aX 9v^2/2a=X 9v^2/2(v^2/2X)=X 9X=X now what?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0X is just for the first situation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[0 = 9v^2 + 2as\] \[9v^2 = 2as\] \[\frac{9v^2}{2a} = s\] \[\frac{9v^2}{2(\frac{v^2}{2X})} = s\] \[\implies \frac{9v^2}{\frac{v^2}{X}} = s\] \[\implies 9X= s\]
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