Question

A particle is moving with velocity v(t) = t2 – 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, that is, s(0) = 1.

The average velocity over the interval 0 to 8 seconds

The instantaneous velocity and speed at time 5 secs

The time interval(s) when the particle is moving right

The time interval(s) when the particle is
going faster
slowing down

Find the total distance the particle has traveled between 0 and 8 seconds

Answer 1

Do you need help with all of these?

Answer 2

well the instantaneous velocity at t=5 can be found by pluggin in t=5
the velocity function is the one given

Answer 3

yes please

Answer 4

so you found the instantaneous velocity yet?

Answer 5

yes which is 88

Answer 6

isn't your equation v=t^2-9t+18

Answer 7

I don't get 88 when pluggin in 5

Answer 8

i rechecked is -2

Answer 9

sounds tons better :

Answer 10

speed=|v|

Answer 11

so is 2

Answer 12

so yeah your second question is actually a whole bunch easier than your first question

Answer 13

because to find the average velocity we need to find the position function

Answer 14

this is because average velocity on [0,8]
equals:
\[\frac{s(8)-s(0)}{8-0}\]

Answer 15

the velocity normally would have been found by finding the derivative of the position function s
but that was given to us
so it made our work for instantaneous rate of change and speed tons easy

Answer 16

do you know how to find s given s'?

Answer 17

2t-t ?

Answer 18

if it wasn't clear from what i was saying
i'm saying velocity=position'
v=s'
so to find s you need to integrate both sides .

Answer 19

\[s'=v \\ \int\limits s' dt=\int\limits v dt \\ s(t)=\int\limits v dt \]

Answer 20

freckles
[Math Processing Error]

Answer 21

have you integrated v yet ?

Answer 22

2t-t? i don't know how to do it !:(

Answer 23

2t-9

Answer 24

8t^2-72t+144 ?

Answer 25

do you have any experience with power rule ?

Answer 26

\[\int\limits t^n dt=\frac{t^{n+1}}{n+1}+C , n \neq -1 \]

Answer 27

the first term you have t^2
use the above formula

Answer 28

you can also use it for the second term -9t

Answer 29

ok

Answer 30

3t^2/3 +c ?

Answer 31

2+1 is 3 not 2 and the 3 in front shouldn't magically appear

Answer 32

\[\int\limits_{}^{}t^2 dt=\frac{t^{2+1}}{2+1}+C=\frac{t^3}{3}+C\]

Answer 33

i thought i have to solve for derivetive

Answer 34

no
like s'=v
so the s=antiderivative of v

Answer 35

@satellite73 I'm going to go if you want to help this person in calculus. I know how much you like calculus. (It is just velocity and speed and all that jazz, nothing too scary).