Question

a particle moves along a line so that its velocity at time t is v(t)= t^2-t-6. ( Measured in meters per second )

a. find the displacement of the particle during the time period 1
9 years ago

Answer 1

@zepdrix

Answer 2

@freckles

Answer 3

@satellite73

Answer 4

first one is the integral

Answer 5

\[\int_1^4 (t^2-t-6)dt\]

Answer 6

you good with that?

Answer 7

yes

Answer 8

second one is not the integral, you have to break it in two parts

Answer 9

does it need absolute value ? for the one you wrote

Answer 10

no

Answer 11

first one is displacement, means how far you were from where you started

Answer 12

second one is total distance including going forward and backward

Answer 13

so then the distance does require the absolute value ?

Answer 14

\[t^2-t-6\] is negative until you get to 3, then positive
that means the particle is going to the left, then to the right

Answer 15

i guess i should ask first if that is clear or no, and did you see how i knew it

Answer 16

im sorry im really confuse with these problems i have trouble what to do

Answer 17

ok lets go slow
first is done though right? just the integral

Answer 18

yes so do we integrate that ?

Answer 19

yes, for the first one
the displacement is \[\int_1^4 (t^2-t-6)dt\] whatever that is
it is an easy enough integral to compute
we can do that second if you like, lets to part two first

Answer 20

okay

Answer 21

\[v(t)=t^2-t-6\] is a parabola that opens up
that is clear, yes?

Answer 22

yes

Answer 23

it also factors nicely as \((t-3)(t+2)\) so it is zero at \(t=-2\) and \(t=3\)

Answer 24

here is the picture
http://www.wolframalpha.com/input/?i=t^2-t-6

Answer 25

you can tell even without the picture that is is negative on \((-2,3)\) and positive on \((3,\infty)\)

Answer 26

of course we are only concerned with the interval \((1,4)\)
on \((1,3)\) it is negative
that means the particle is moving to the LEFT
on \((3,4)\) \(v\) is positive, that means it is moving to the RIGHT

Answer 27

so far so good?

Answer 28

oh okay is there a way to do it algebraically to know that ?

Answer 29

it is pretty obvious right? a parabola that opens up is negative between the zeros and positive outside them

Answer 30

so you have to break the integral in to two parts \[\int_1^3v(t)dt+\int _3^4v(t)dt\] the first one will be negative, make it positive

Answer 31

yes do its just integrates and itll be f(b)- f(a)

Answer 32

which one we doing now first or second?

Answer 33

part A is displacement you don't split them
part B is the distance travelled, that is the one you split

Answer 34

oh okay a. is -4.5 and now we doing b lol

Answer 35

you want me to check?

Answer 36

how would be look like ?