Question

A particle moves on the x-axis so that its position at any time is given by x(t)=.01t^4-.04t^3.


c. Find the total distance traveled by the particle from t=0 to t=5.
A particle moves on the x-axis so that its position at any time is given by x(t)=.01t^4-.04t^3.


c. Find the total distance traveled by the particle from t=0 to t=5.
@Mathematics

Answer 1

just integrate x(t) from t=0 to t=5!

Answer 2

i could be wrong but i don't think that is going to work because this function is not positive

Answer 3

1.25m just plug in the five where you see t

Answer 4

the answer is zero!

Answer 5

this is negative from 0 to 4, then positive.

Answer 6

ok now i know that is not right. it is asking for the total distance traveled.

Answer 7

the position is given in function of t!!!! how is it going to be zero?

Answer 8

it isn't. you go to the left, then to the right.

Answer 9

right so the total distance would be 1.79 plugin the values of t

Answer 10

i think you're right it just seems too easy.

Answer 11

yes, the first three seconds goes to the left 0.27 one second ater that goes from left to right 0.27 thats 0.54 and one second after that goes rom left to right 1.25 adding all up is 1.79

Answer 12

ok i am tired and i may be off, but i think you have to integrate over where it is positive and where it is negative
in other words take
\[-\int_0^4x(t)dt+\int_4^5 x(t) dt\]

Answer 13

oh wait that is wrong you have the position function, not the velocity

Answer 14

i think satellite73 is correct

Answer 15

ok lets do it the correct way. you start at position 0

Answer 16

not wrong! you are integrating the position.... you will have the distance!

Answer 17

you are going to the left. i know that because this is
\[x(t) = (0.01 t-0.04) t^3\]

Answer 18

now let us see how far we go to the left. if you take the derivative you get
\[x'(t)=0.04 x^3-0.12 x^2= 0.04 (x-3) x^2\]

Answer 19

so your function is decreasing (your particle is going to the left) from t = 0 when it is at 0 to t = 3

Answer 20

now at t = 3 we are at
\[x(3)= -.27\] and that is how far you have gone to the left.

Answer 21

then you head to the right and at t = 5 you get
\[x(5)=1.25\] or a total distance of just what santi said.

.27 + .27 +1.25

Answer 22

ok thank you