Question

Describe the motion of the particle for t is greater than or equal to 0. x(t)=t^2-4t+3. x(t) is the position function

Answer 1

@jim_thompson5910

Answer 2

why not graph it and describe that? That's what I would do

Answer 3

Im not allowed to use my calculator

Answer 4

Lol i wish I could

Answer 5

ok, the next best thing is to graph by hand

Answer 6

so create a table to help you plot a bunch of points

then draw a curve through them all

Answer 7

Ok, so I plug in values for t until I find the end behavior? It has something to do with concavity and extrema because thats what we've been doing

Answer 8

ok, so it looks like they want you to find the critical point

Answer 9

what is x ' (t) ?

Answer 10

2t-4

Answer 11

set that equal to zero to get t = ??

Answer 12

oh ok

Answer 13

So then I make a sign chart?

Answer 14

yes

Answer 15

Ok

Answer 16

Thanks!!

Answer 17

np

Answer 18

Wait, but this says that it is moving left when t<2. Why?

Answer 19

http://www.slader.com/textbook/9780132014083-calculus-graphical-numerical-algebraic-3rd-edition/215/exercises/25c/

Answer 20

it moves left when the velocity is negative

the velocity is negative when x ' (t) < 0

Answer 21

x(t) is the position function (only along the x axis)

x ' (t) is the velocity function

Answer 22

if you graph x ' (t) = 2t - 4, you'll get this

Answer 23

But when I did the sign chart thing, I got - and then +

Answer 24

That just means that the slope of the position graph is negative and then positive

Answer 25

the portion of the derivative that is below the horizontal axis refers to when x ' (t) is negative