Find the total distance traveled during the first 8s for v = 3t^2 - 24t + 36 -- just tell me what steps i'm supposed to follow to to this, i know the particle is at rest at t = 2,6 and moving in the positive direction when 0=6

Question

Find the total distance traveled during the first 8s for v = 3t^2 - 24t + 36 -- just tell me what steps i'm supposed to follow to to this, i know the particle is at rest at t = 2,6 and moving in the positive direction when 0=<t<2 and t>6

campbell_st · Answer

is V a velocity equation...?

anonymous · Answer

Yes that's correct

campbell_st · Answer

well you need to integrate the velocity equation to obtain the displacement equation

then you will need to find the value of the constant.

anonymous · Answer

i dont think we really need to find the constant since he want distances..

anonymous · Answer

take mod || of the velocity put dt in the end and integerate it from 0 to 8 
u will have to break the limit as its distance not displacement  at  2 and 6 
and put it negative from 2 to 6 and the rest should be positive....

anonymous · Answer

the way I was understanding it was that I was just needing to figure out the displacements between the positive and negative directions and add them up

anonymous · Answer

alright let me try that

anonymous · Answer

$$v = 3t^2 - 24t + 36 $$

Just u have to Do is that Intergrate...Both side

anonymous · Answer

@Yahoo!  this will give displacement not distance..

anonymous · Answer

Oh....Sorry..i Did nt read the question well

anonymous · Answer

$$|v| = s$$

$$S = |  3t^2 - 24t + 36 |$$

anonymous · Answer

ok no prob..  bt s should be speed here .. :P