A community for students.
Here's the question you clicked on:
 0 viewing
Callisto
 4 years ago
Deriving kinematic equations
Callisto
 4 years ago
Deriving kinematic equations

This Question is Closed

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3\[v=\int a dt=at+c\]c = initial velocity So, \[v=u+at\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if a constan, a =0 v = u + at v = u + 0 v = u

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3\[x_f = \int v(t) dt=\int (\int a dt) dt = \int(at+u)dt = \frac{1}{2}at^2 + ut +c\], c = initial position So, \[x_f = ut +\frac{1}{2}at^2 + x_i\]\[x_fx_i = ut +\frac{1}{2}at^2\]\[s = ut+\frac{1}{2}at^2\]s=displacement.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for a = 0; s = ut + 0.5 at^2 s = ut + 0 s = ut

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3From the third post, \[s=ut+\frac{1}{2}at^2\]\[at^2+2ut 2s = 0\]\[t^2 + \frac{2u}{a}t  \frac{2s}{a}=0\]\[(t+\frac{u}{a})^2  \frac{u^2}{a^2}\frac{2s}{a}=0\]\[(t+\frac{u}{a})^2 = \frac{u^22as}{a^2}\]\[t+\frac{u}{a} = \frac{\sqrt{u^22as}}{a}\]\[t = \frac{\sqrt{u^22as}u}{a} (1)\] Put (1) into v=u+at \[v =u+a(\frac{\sqrt{u^22as}u}{a})\]\[v = u +\sqrt{u^22as}u\]\[v^2 = u^2 +2as\]

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3\[x_f = \int v(t) dt = v_{ave}t+C\]c = initial position \[x_f = v_{ave}t +x_i\]\[x_f = \frac{1}{2}(u+v)t +x_i \]\[s = \frac{1}{2}(u+v)t\]displacement = s = \(x_f  x_i\)

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3Probably something wrong with the last post, which is s=(1/2) (u+v)t.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0that's correct ... it assumes constant acceleration. that;s all.

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3Seriously?! I did it!?!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0let me see how can i put it up logically.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351833416633:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0now you just need to show that for constant accn V_av = (u+v)/2 ... wanna try it?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3Huh!? I thought it's some maths.. Oh..How to start??

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0well, you can do this without MVT ,,, i was wondering if i could improve my skills with MVT. try expanding v(t) in the expression of average.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0or simply put ,,, average velocity = distance/time

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3\[v_f= v_i+\int_0^{t_f} a dt=v_i + at\] So, \[v_f=v_i + at\]

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3\[x_f = x_i+\int_0^{t_f} v(t) dt= x_i+\int_0^{t_f} (v_i+at) dt = x_i+v_it+\frac{1}{2}at^2\] So, \[x_f = x_i+v_it+\frac{1}{2}at^2\] That is \[x_f  x_i = v_it+\frac{1}{2}at^2\]\[s= ut+\frac{1}{2}at^2\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351853269560:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351853406639:dw
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.