## anonymous one year ago 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

1. freckles

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

2. freckles

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

3. anonymous

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

4. anonymous

would the average velocity be 3.33

5. phi

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

6. freckles

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

7. freckles

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

8. anonymous

instantantaneous velocity is dx/dt

9. anonymous

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

10. freckles

yep

11. freckles

are you sure @phi that is what I got

12. anonymous

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

13. freckles
14. phi

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

15. anonymous

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

16. anonymous

thank you very much!

17. phi

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