Question

The velocity function (in meters per second) is given for a particle moving along a line. Find a) the displacement and b) the distance traveled by the particle during the given time interval?
V(t)=t^2-2t-8 1 10 years ago

Answer 1

@zepdrix

Answer 2

The velocity function is the `derivative` of the displacement function.
So then, we can find the displacement function by `anti-differentiation` of the velocity function, ya?

Answer 3

Yes

Answer 4

\[\large\rm s(t)=\int\limits_1^6 v(t)dt\]

Answer 5

\[\large\rm s(t)=\int\limits_1^6 t^2-2t-8~dt\]

Answer 6

Apply `power rule for integration` to each term, ya? :)

Answer 7

Yes :O

Answer 8

Do you want me to do it? :O

Answer 9

ya sure

Answer 10

Okay

Answer 11

1/3t^3-t^2-5t?

Answer 12

:o

Answer 13

1/3t^3-t^2-8t, ya

Answer 14

Oops :D

Answer 15

Then evaluate at the limits,\[\large\rm \left(\frac{1}{3}t^3-t^2-8t\right)_{t=6}-\left(\frac{1}{3}t^3-t^2-8t\right)_{t=1}\]

Answer 16

\[\large\rm \left(\frac{1}{3}6^3-6^2-8(6)\right)-\left(\frac{1}{3}1^3-1^2-8(1)\right)\]

Answer 17

Woooooo :O

Answer 18

Okay just give me a second to figure this out

Answer 19

-10/3

Answer 20

@zepdrix

Answer 21

yay good job

Answer 22

Now I have to do the other part :(

Answer 23

@zepdrix we have to do the next part!!!!

Answer 24

soz

Answer 25

\[\large\rm v(t)=t^2-2t-8\]We need to figure out when the velocity is zero,
that tells us at which points the particle is changing directions.

Answer 26

\[\large\rm 0=t^2-2t-8\]factor

Answer 27

(t-4)(t+2)

Answer 28

Ok at time t=4 our velocity is zero,
that is where it changes direction.

Answer 29

Oh you have to show that?

Answer 30

So we have to do something fancy.
We have to break up our integral.\[\large\rm \int\limits_1^6\quad=\quad \int\limits_1^4\quad+\quad\int\limits_4^6\]

Answer 31

Ugh I don't get doing that

Answer 32

\[\large\rm d(t)=\left|\int\limits_1^4 t^2-2t-8~dt\right|+\left|\int\limits_4^6 t^2-2t-8~dt\right|\]

Answer 33

no?

Answer 34

here is an example