Rates of Change on Vert Line: Particle s(t)

[tagrico]

edit: meant to put rates of change as the title, not related rates.

Homework Statement

A particle moves on a vertical line so that its coordinates at time t is s(t) = 4t^3 - 12t^2 +7, where t≥0
a) when is the particle moving upward and when is it moving downward
b) find the total distance that the particle travels in the time interval 0≤t≤ 3

The Attempt at a Solution

so I'm pretty sure I would take the derivative to find the velocity function, but how do I find out when it's moving upward and when it is moving downward? positive = upward and negative = downward?

 

[Dick]

Sure, they are sort of implying since the motion is along a vertical line that s(t) is distance along that line upward from the origin. So positive derivative is upwards.

 

[tagrico]

When I take the derivative I get 12t^2 -24t. Are you saying this can only be a positive function?

 

[Dick]

When I take the derivative I get 12t^2 -24t. Are you saying this can only be a positive function?

No! I'm saying for values of t where the derivative is positive motion is upwards, for values of t where the derivative is negative, it's downwards. The question is when (for what values of t) is it upwards and for which is it downwards.

 

[tagrico]

Gotcha! I'm stuck on which values of t is would be upward and downward. Other than plugging in random values of t into the velocity function, how would I go about doing this?

 

[HallsofIvy]

Factor it. $12t^2- 24t= 12t(t- 2)$. That will be positive when the two variable factors, t and t-2, have the same sign, negative when they have different signs.

Also, "a- b" is positive if and only if a> b, negative if and only if a< b so the signs of t- 2 and t= t- 0 depende upon whether t> 2 and t> 0. If t< 0, then t< 2 also so t and t- 2 are both negative. If 0< t< 2, t is positive and t- 2 is negative. If t> 2 then t> 0 also so t and t- 2 are both positive.