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.

1.The average velocity over the interval 0 to 8 seconds

2.The instantaneous velocity and speed at time 5 secs
Help please

Answer 1

you can find the position function by integrating the velocity function and then apply the condition s(0)=1

Answer 2

you can find the average velocity over the interval t in [0,8] by computing:
\[\frac{s(8)-s(0)}{8-0}\]

Answer 3

acceleration ->integrate -> velocity
velocity -> integrate -> displacement

Answer 4

would the average velocity be 3.33

Answer 5

oh wait. first subtract off s(0)= 1

Answer 6

can you find the instantaneous velocity? (hint: think derivative of position)

Answer 7

and you are actually already given the derivative of the position :)

Answer 8

instantantaneous velocity is dx/dt

Answer 9

so you plug in 5 into the v(t)?

Answer 10

yep

Answer 11

are you sure @phi
that is what I got

Answer 12

how do I find the average velocity with initial condition s(0)=1

Answer 13

http://www.wolframalpha.com/input/?i=1%2F8*%28%288%5E3%2F3-9%2F2%288%29%5E2%2B18%288%29%2B1%29-%281%29%29

Answer 14

oh, never mind. that constant is subtracted off.
so yes 3.333

Answer 15

you can always check your answer on wolframalpha, but phi has a good explanation on it

Answer 16

thank you very much!

Answer 17

how do I find the average velocity with initial condition s(0)=1

after integrating v(t) you get
\[ \frac{t^3}{3}- \frac{9t^2}{2} +18t + C \]
when t is 0 that should be 1, i.e. C= 1
and when we do s(8)- s(0) the one's cancel.
so we just need
\[ \frac{t^3}{3}- \frac{9t^2}{2} +18t\]
evaluated at t= 8 (for the top)
the bottom is 8-0= 8