help please!!!!

- YumYum247

help please!!!!

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- YumYum247

##### 1 Attachment

- YumYum247

@Elsa213 give it a shot!!!!

- Teddyiswatshecallsme

I don't fluttering know.
;~;

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## More answers

- YumYum247

awn :(

- Teddyiswatshecallsme

Yeah. Sorry dude.

- YumYum247

dude? how do u know my gender??????????

- YumYum247

@oldrin.bataku plz help :)

- YumYum247

@oldrin.bataku help plz

- arindameducationusc

Let me try....

- arindameducationusc

Do I have to check the steps?

- YumYum247

sure i guess!!! O-o

- 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

yesh

- anonymous

oops, actually i forgot to divide by 60 again for 1 min = 60 s

- anonymous

so \(v_i\approx20.83\) m/s

- YumYum247

yah i just nticed that too :D

- 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

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

so am i right?????????????? XD

- arindameducationusc

I don't get it. why average velocity is the constant deacceleration? It is not necessary
@oldrin.bataku

- arindameducationusc

It should be.... x/t=v+u/2 ...
for x

- 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

in fact, none of our kinematic equations we hope to use will work in the case the acceleration is not constant

- arindameducationusc

That is only Yumyum did....
Ofcourse it wouldn't work. It should be given in question though....

- YumYum247

i wrote down the question as is......

- arindameducationusc

@oldrin.bataku What did you say in my question of Big O IO did not understand......

- YumYum247

@arindameducationusc let's just postpone for now......

- arindameducationusc

okay

- YumYum247

i don't want my master to get his fingers tired XD

- YumYum247

Thank you @oldrin.bataku :")

- 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

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

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

if the acceleration is constant, yes, since then the velocity vs time graph is a straight line and that's just the midpoint

- YumYum247

ok thanks :"|

- arindameducationusc

@oldrin.bataku Nice... but can you explain in terms of graph? By making a graph in rough @oldrin.bataku

- anonymous

http://cs.anu.edu.au/~Alistair.Rendell/Teaching/apac_comp3600/module1/images/Introduction_BigOGraph.png

- arindameducationusc

Okay thanks @oldrin.bataku

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