Question

Find the acceleration of this particle at all time values where this particle stops

Answer 1

This should be in physics, but can I have the values please (:

Answer 2

A particle is moving along the x-axis. At time t, the particle is at x(t)= 2t - 9t^2 +12t +2. Find the acceleration of this particle at all time values where this particle stops.

Answer 3

I guess you can start by finding the roots of the equation to get you the graph of the function. If the particles stops, there is no acceleration and velocity = 0.

Answer 4

Hmmmm. We are going over derivatives and differentiation in class, if that makes it make any more sense.

Answer 5

hmm this must be calculus :P
I dont have it till next semester. but ill learn what I can (:

Answer 6

as of now

Answer 7

lol, it is indeed calculus, and a pain in my butt! :p

Answer 8

lool you must be a senior, looking at these online notes make my head spin. It is pretty confusing eh?

Answer 9

This is what I'm looking at: http://mathworld.wolfram.com/Derivative.html

Answer 10

that comic strip at the end of the page :P

Answer 11

hahaha the comic strip was funny and sooooo true!

and actually I'm a junior in college hahaha.

Answer 12

ahah yeah :D and aha oh really? that's neat!
Well i'll call for help, see if anyone sees this message, if only i knew this stuff :(

Answer 13

thanks! I'm meeting a friend now to see if he can help me out! thanks for trying lol.

Answer 14

lol alrii, it's night time here, so ill just meet my bedd...
Night ;P

Answer 15

@sauravshakya @ujjwal @timo86m @mukushla @TuringTest
Not my question, but hai if anyone of you could solve it, that would be great

Answer 16

find x'(t)

Answer 17

how would you do that? find roots?

Answer 18

no

Answer 19

then?

Answer 20

I think its x(t)= 2t^3 - 9t^2 +12t +2.
Now, velocity=x'(t)=6t^2 -18t+12
And accleration=x''(t)=12t
Now, when the particle stops ,
velocity=0
6t^2 -18t+12=0
t^2-3t+2=0
t^2 -2t -t +2=0
t(t-2)-1(t-2)=0
(t-1) (t-2)=0
Thus, t=1 or 2
Now,
when t=1 , accleration=12*1=12
when t=2 then , accleration =12*2=24

Answer 21

does it help?