SOLUTION: A particle moves on a vertical line so that it's coordinate in feet at time t is s(t) = 4t^3 - 12t^2 + 7 , t ≥ 0 a) Find the velocity and acceleration functions. b)

Question 911874: A particle moves on a vertical line so that it's coordinate in feet at time t is s(t) = 4t^3 - 12t^2 + 7 , t ≥ 0
a)
Find the velocity and acceleration functions.
b)
When is the particle moving upward and when is it moving downward?
c
Find the total distance that the particle travels in the time interval 0 ≤ t ≤ 3

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A particle moves on a vertical line so that it's coordinate in feet at time t is s(t) = 4t^3 - 12t^2 + 7 , t >= 0
a)
Find the velocity and acceleration functions.
s(t) = 4t^3 - 12t^2 + 7
Velocity is s' = 12t^2 - 24t
Accel is s" = 24t - 24
--------------------
b)
When is the particle moving upward and when is it moving downward?
Vel + --> up
12t^ - 24t > 0
Solve 12t^2 - 24t = 0 --> Vel + for t > 2
-----------------
Vel - --> down
12t^ - 24t < 0
0 <= t <= 2
-------------------------
c
Find the total distance that the particle travels in the time interval 0 <= t <= 3
============
Distance = 32.292929...
I'll do that later.

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SOLUTION: A particle moves on a vertical line so that it's coordinate in feet at time t is s(t) = 4t^3 - 12t^2 + 7 , t &#8805; 0 a) Find the velocity and acceleration functions. b)

SOLUTION: A particle moves on a vertical line so that it's coordinate in feet at time t is s(t) = 4t^3 - 12t^2 + 7 , t ≥ 0 a) Find the velocity and acceleration functions. b)