Question

help please!!!!

Answer 1

@Elsa213 give it a shot!!!!

Answer 2

I don't fluttering know.
;~;

Answer 3

awn :(

Answer 4

Yeah. Sorry dude.

Answer 5

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

Answer 6

@oldrin.bataku plz help :)

Answer 7

@oldrin.bataku help plz

Answer 8

Let me try....

Answer 9

Do I have to check the steps?

Answer 10

sure i guess!!! O-o

Answer 11

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

Answer 12

yesh

Answer 13

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

Answer 14

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

Answer 15

yah i just nticed that too :D

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

@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)

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

okay

Answer 28

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

Answer 29

Thank you @oldrin.bataku :")

Answer 30

@arindameducationusc we say that \(f(x)\in O(g(x))\) if we mean that eventually (i.e. for sufficiently big \(x\)) we have that \(|f|<C|g|\), so the magnitude is bounded above by some constant \(C\) times the magnitude of \(g\)

Answer 31

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

Answer 32

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),

Answer 33

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

Answer 34

ok thanks :"|

Answer 35

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

Answer 36

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

Answer 37

Okay thanks @oldrin.bataku