A community for students.
Here's the question you clicked on:
 0 viewing
YumYum247
 one year ago
help please!!!!
YumYum247
 one year ago
help please!!!!

This Question is Closed

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1@Elsa213 give it a shot!!!!

Teddyiswatshecallsme
 one year ago
Best ResponseYou've already chosen the best response.0I don't fluttering know. ;~;

Teddyiswatshecallsme
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. Sorry dude.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1dude? how do u know my gender??????????

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1@oldrin.bataku plz help :)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1@oldrin.bataku help plz

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Let me try....

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Do I have to check the steps?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, actually i forgot to divide by 60 again for 1 min = 60 s

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \(v_i\approx20.83\) m/s

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1yah i just nticed that too :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0supposing 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.1so am i right?????????????? XD

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0I don't get it. why average velocity is the constant deacceleration? It is not necessary @oldrin.bataku

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0It should be.... x/t=v+u/2 ... for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0in fact, none of our kinematic equations we hope to use will work in the case the acceleration is not constant

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0That is only Yumyum did.... Ofcourse it wouldn't work. It should be given in question though....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1i wrote down the question as is......

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku What did you say in my question of Big O IO did not understand......

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1@arindameducationusc let's just postpone for now......

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1i don't want my master to get his fingers tired XD

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1Thank you @oldrin.bataku :")

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@arindameducationusc we say that \(f(x)\in O(g(x))\) if we mean that eventually (i.e. for sufficiently big \(x\)) we have that \(f<Cg\), so the magnitude is bounded above by some constant \(C\) times the magnitude of \(g\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which 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 xaxis

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1hey 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
 one year ago
Best ResponseYou've already chosen the best response.0if the acceleration is constant, yes, since then the velocity vs time graph is a straight line and that's just the midpoint

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku Nice... but can you explain in terms of graph? By making a graph in rough @oldrin.bataku

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Okay thanks @oldrin.bataku
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.