Question

1) A particle is moving with velocity:
v(t) = t^2 -9t+18

with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero.

Answer 1

\[v(t) = t^{2} - 9t + 18\]

Answer 2

so what is the question?

Answer 3

Find each of the following. Give units in the answer.

a) The initial acceleration
b) The average velocity over the interval 0 to 8 seconds
c) The instantaneous velocity at time 5 secs
d) The time(s) when the particle is at rest
e) The time interval(s) when the particle is moving left
f) The time interval(s) when the particle is moving right
g) The speed of the particle at time 4 secs
h) The time interval(s) when the particle is
(i) going faster
(ii) slowing down
i) Find the total distance the particle has traveled between 0 and 8 seconds

Answer 4

@sourwing srry, forgot to post it.

Answer 5

O.O

Answer 6

this question is annoying me T_T

Answer 7

Which part are you having trouble with?

Answer 8

i dont know how to find initial accel

Answer 9

Acceleration is the derivative. So \(a_0 = v'(0)\)

Answer 10

im guessing i take the derivative, but then, what do they mean by initial.

Answer 11

ah. ty.

Answer 12

and what about average velocity?

Answer 13

Given no information, we assume \(t_i=0\) and so \(a_i=a(t_i)=v'(0)\)

Answer 14

Average velocity done here by using calculating the average of the function \(v\) over the interval from \(0\) to \(8\).

Answer 15

so i take the velocity of each, and divide by 8, or is there an easier way?

Answer 16

Do you know how to calculate the average value of a function over an interval?

Answer 17

i forgot tbh.

Answer 18

\[
\overline{f_{a\to b}} = \frac{1}{b-a}\int\limits_a^bf(t)\;dt = \frac{F(b)-F(a)}{b-a}
\]Where \(F' =f\)

Answer 19

o, u mean the mean value theorem, so c would be the average velocity?

Answer 20

or in this case, f

Answer 21

Welll, in terms of the mean value theorem, in this case \(c\) is the time at which \(v\) is at it's average value on the \([a,b]\) interval.

Answer 22

aite, ty, i should be able to handle the rest, can i message you if i get stuck again?

Answer 23

Mean value theorem only tells us that at some point in our interval we reach the mean value.

What I gave you resembles the mean value theorem, but its purpose is to calculate the mean value.

Answer 24

If you get stuck, let me know.

Answer 25

ty

Answer 26

and what is the distance the particle has traveled.

Answer 27

when s(t) is positive, is it going to the left or right?

Answer 28

is right positive, or is left positive?

Answer 29

it doesnt say, im guessing left is negative tho, so how do i find the distance tho?

Answer 30

Use the following:\[
\int _0^tv(t)\;dt = s(t)-s(0) = s(t)-1
\]Assuming that positive means right.

Answer 31

tyvm