Question

A particle moves along the x-axis so that at any time 0 11 years ago

Answer 1

I have to use a graphing calculator, but I am still unsure of how to do particle motion problems.

Answer 2

\[\large{v(t) = (t-2)^2 \cos(2t)}\]

The particle will change its direction everytime sign of v changes

Answer 3

http://www.wolframalpha.com/input/?i=graph+%28t-2%29%5E2*cos%282t%29+++from+0%3Ct%3C5

Answer 4

Now I never used graphing calculator. So, you will have to tell me this:

Does graphing calculator draw a graph ?

A stupid question though

Answer 5

*My question is stupid

Answer 6

yes, it will and it will also draw derivatives

Answer 7

Okay so draw the graph and see the signs

Answer 8

your question is fine, I am just well verse on the graphing calculator

Answer 9

I typed it into wolframalpha if you want to look at the graph

Answer 10

Okay I am going to use wolfram to plot the graph

Answer 11

I want to say 3 times

Answer 12

+ -> - -> rest -> + -> -

Answer 13

I am not sure whether rest would affect the answer or not

Answer 14

yes I am not sure about the "rest" either. I never studied physics

Answer 15

Okay lets see in physical conditions.

Answer 16

well my next question leads me into v ' (2) and to describe the motion
so my answer would be v ' (2)=0 and the particle is at rest

Answer 17

\[\large{v'(t) = 2(t-2)\cos(2t) - 2(t-2)^2\sin(2t)}\]

Yeah v'(2) = 0

Answer 18

wow you are good. I took the easy way out and just let the calculator solve it. and yes the calculator can find derivatives at a point

Answer 19

Thanks :)

Answer 20

last question, average acceleration, do I take the integral for that?

Answer 21

Yeah

Answer 22

\[\frac{ 1 }{ b-a}\int\limits_{a}^{b}v(t)dt\]

Answer 23

Yeah

Answer 24

or is it the function a(t)dt?

Answer 25

I don't think there is much difference

Answer 26

ok let me calculate both since I can do it on my calculator

Answer 27

Yeah that would do :)

Answer 28

.2819 is for v(t)dt
.84496 is for a(t)dt

Answer 29

I really don't like particle motion

Answer 30

Here! here! :D

Answer 31

Okay figured out a(t) dt is correct

Answer 32

Integral gives area

Answer 33

Area of v(t) and t curve is total displacement and thus will give average velocity

Answer 34

Area of a(t) and t curve would give net change in velocity and thus average acceleration

Answer 35

ok thanks, I went to google it and got even more confused

Answer 36

Google's one of few disadvantages :)

Answer 37

true

Answer 38

thank you so much, I hate particle motion. Maybe I should go do over type of homework instead and skip this nonsense. I wish I had taken physics at some point because it really gives me a lot of heartache.

Answer 39

Your welcome :)