Question

Need some assistance if anyone is available:

Problem #7:
At time t in seconds, a particle's distance s(t), in micrometers (um), from a point is given by s(t)=e^(t)-1. What is the average velocity of the particle from t=2 to t=4?

If anyone has some time right now, i'd appreciate it if someone could work through this problem with me.

Answer 1

ok first find s at time t = 2 seconds
then at time t = 4 secs

then subtract the distances s(4) - s(2)
and divide by 4-2 = 2 seconds

average velocity = (s(4) - s(2)) / 2

Answer 2

just to clear things up, to find s at time t=2 i plug t=2 into s(t)=x^(t)-1 correct?

Answer 3

yes

Answer 4

alright let me try all that out.

Answer 5

you should get 6.389 for that

Answer 6

oooh ok i see what i did before, before i was using ln() to get rid of the e and to bring the exponent of 2 down, i thought thats what i had to do because of some stuff i was doing earlier that required that method (think derivatives)

Answer 7

just plug t = 2 into the formula

Answer 8

ok so for
t=2 you get 6.389
t=4 you get 53.598

Answer 9

so next is \[\frac{ 53.589-6.389 }{ 2 }\]

Answer 10

giving me 23.6045

Answer 11

so my avg velocity is 23.6 um/s

Answer 12

i got 23.796

Answer 13

hm let me try again

Answer 14

no sorry you are right

Answer 15

hm not sure how

Answer 16

oh haha alright :D

Answer 17

23.6 is correct

Answer 18

awesome so that is my avg velocity right?

Answer 19

yes

Answer 20

oh wow, i was so close the first time, i just added a weird step that ruined the rest

Answer 21

the units are um per second or um s(-1

Answer 22

or as you wrote it um/s

Answer 23

yeah i got the unit part, i just didnt write it that last time

Answer 24

ok - average velocity = total distance / time taken

Answer 25

gotta go - good luck

Answer 26

alright thanks for the help!