Question

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, 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

The instantaneous velocity and speed at time 5 secs is v(t)=\[t=5 , v(t)=t^2-9t+18 =(5)^29(5)+18= 25-45+18=/-2/=2\]

Answer 2

*The average velocity over the interval 0 to 8 seconds*
work out how far it actually travels in 8 sec and divide by 8.

integration: bearing in mind, "time t = 0 sec is 1 meter right of zero"

Answer 3

v(t) = t^2 – 9t + 18
average velocity = (t^2-9t+18)/8=(8^2-9*8+18)/8
instantaneous velocity v(5)=5^2-9*5+18
you could plot the funtion V vs. t

Answer 4

@AJ01 what do you mean V vs. t? i am sorry i have been out off school for a week don't remember anything from that class.

Answer 5

Like he wants you to graph v(t)=t^2-9t+18

Answer 6

V: velocity vs. time

Answer 7

since you are finding total distance, you want to integrate the absolute value of velocity

Answer 8

integral |v(t) |

Answer 9

@perl /V(t)/=2

Answer 10

total distance should be 107/3

Answer 11

35.67

Answer 12

are you allowed to use a calculator?

Answer 13

do you have a TI 83

Answer 14

@perl yes, but i don't have it with me now.

Answer 15

Math->9:fnInt(

fnInt(abs(x^2-9*x+18, x , 0, 8 ) ) = 35.666666666... = 107/3

Answer 16

when you get a chance take out your calculator

Answer 17

@perl thank you soooooooo much.