## YumYum247 one year ago help please!!!!

1. YumYum247

2. YumYum247

@Elsa213 give it a shot!!!!

3. Teddyiswatshecallsme

I don't fluttering know. ;~;

4. YumYum247

awn :(

5. Teddyiswatshecallsme

Yeah. Sorry dude.

6. YumYum247

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

7. YumYum247

@oldrin.bataku plz help :)

8. YumYum247

@oldrin.bataku help plz

9. arindameducationusc

Let me try....

10. arindameducationusc

Do I have to check the steps?

11. YumYum247

sure i guess!!! O-o

12. 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

13. YumYum247

yesh

14. anonymous

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

15. anonymous

so $$v_i\approx20.83$$ m/s

16. YumYum247

yah i just nticed that too :D

17. 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

18. 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

19. YumYum247

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

20. arindameducationusc

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

21. arindameducationusc

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

22. 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)

23. anonymous

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

24. arindameducationusc

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

25. YumYum247

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

26. arindameducationusc

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

27. YumYum247

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

28. arindameducationusc

okay

29. YumYum247

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

30. YumYum247

Thank you @oldrin.bataku :")

31. 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 $$|f|<C|g|$$, so the magnitude is bounded above by some constant $$C$$ times the magnitude of $$g$$

32. 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

33. 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),

34. anonymous

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

35. YumYum247

ok thanks :"|

36. arindameducationusc

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

37. anonymous
38. arindameducationusc

Okay thanks @oldrin.bataku