YumYum247
  • YumYum247
help please!!!!
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
YumYum247
  • YumYum247
YumYum247
  • YumYum247
@Elsa213 give it a shot!!!!
Teddyiswatshecallsme
  • Teddyiswatshecallsme
I don't fluttering know. ;~;

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YumYum247
  • YumYum247
awn :(
Teddyiswatshecallsme
  • Teddyiswatshecallsme
Yeah. Sorry dude.
YumYum247
  • YumYum247
dude? how do u know my gender??????????
YumYum247
  • YumYum247
@oldrin.bataku plz help :)
YumYum247
  • YumYum247
@oldrin.bataku help plz
arindameducationusc
  • arindameducationusc
Let me try....
arindameducationusc
  • arindameducationusc
Do I have to check the steps?
YumYum247
  • YumYum247
sure i guess!!! O-o
anonymous
  • anonymous
so he's going \(v_i=75\text{ km/h}=75000\text{ m/h}=1250\text{ m/s}\) and \(v_f=0\) m/s over a time span of \(3.4\) s
YumYum247
  • YumYum247
yesh
anonymous
  • anonymous
oops, actually i forgot to divide by 60 again for 1 min = 60 s
anonymous
  • anonymous
so \(v_i\approx20.83\) m/s
YumYum247
  • YumYum247
yah i just nticed that too :D
anonymous
  • anonymous
supposing the brakign is a constant deceleration, the average velocity during breaking is \(\bar v=\frac12(v_i+v_f)\approx \frac12 (0+20.83)=10.42\) m/s
anonymous
  • anonymous
so in \(3.4\) s the distance the car goes through is \(\bar v\cdot \Delta t\approx10.42\cdot 3.4=35.428\) m
YumYum247
  • YumYum247
so am i right?????????????? XD
arindameducationusc
  • arindameducationusc
I don't get it. why average velocity is the constant deacceleration? It is not necessary @oldrin.bataku
arindameducationusc
  • arindameducationusc
It should be.... x/t=v+u/2 ... for x
anonymous
  • anonymous
@arindameducationusc if the braking is not a constant deceleration then we cannot reason the average velocity is \(\bar v=\frac12 (v_i+v_f)\), since this is the midpoint formula and only applies where the velocity is varying with a constant slope (i.e. \(v=v_0+at\) where acceleration \(a\) is constant)
anonymous
  • anonymous
in fact, none of our kinematic equations we hope to use will work in the case the acceleration is not constant
arindameducationusc
  • arindameducationusc
That is only Yumyum did.... Ofcourse it wouldn't work. It should be given in question though....
YumYum247
  • YumYum247
i wrote down the question as is......
arindameducationusc
  • arindameducationusc
@oldrin.bataku What did you say in my question of Big O IO did not understand......
YumYum247
  • YumYum247
@arindameducationusc let's just postpone for now......
arindameducationusc
  • arindameducationusc
okay
YumYum247
  • YumYum247
i don't want my master to get his fingers tired XD
YumYum247
  • YumYum247
Thank you @oldrin.bataku :")
anonymous
  • anonymous
@arindameducationusc we say that \(f(x)\in O(g(x))\) if we mean that eventually (i.e. for sufficiently big \(x\)) we have that \(|f|
anonymous
  • anonymous
which is why it's called asymptotic -- "in the long run"; we're looking at the comparative behavior of the functions for larger and larger \(x\), further and further along the x-axis
YumYum247
  • YumYum247
hey one last thing, so this is how you get the average velocity if you have the initial and final velocity????? v¯=1/2(vi+vf),
anonymous
  • anonymous
if the acceleration is constant, yes, since then the velocity vs time graph is a straight line and that's just the midpoint
YumYum247
  • YumYum247
ok thanks :"|
arindameducationusc
  • arindameducationusc
@oldrin.bataku Nice... but can you explain in terms of graph? By making a graph in rough @oldrin.bataku
anonymous
  • anonymous
http://cs.anu.edu.au/~Alistair.Rendell/Teaching/apac_comp3600/module1/images/Introduction_BigOGraph.png
arindameducationusc
  • arindameducationusc
Okay thanks @oldrin.bataku

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