# How would I find the open intervals where the particle is slowing down and speeding up?? My work is shown down below.?

Suppose that a particle moves along a horizontal coordinate line in such a way that its position is described by the function s(t)=4t^3-26t^2+56t+2

1. The particles velocity is v(t)=12t^2-52t+56

2. Moving right on: (0,2),(7/3,4)

Moving left on: (2,7/3)

3. The particles acceleration is a(t)=24t-52

4. I got this part wrong..

Determine the open intervals on which the particle is speeding up and slowing down

Slowing down on: (0,13/6) wrong answer

Speeding up on: (0,13/6) wrong answer

I took the acceleration which was 24t-52 set it equal to 0 to get t=52/24=13/6

### 1 Answer

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- stanschimLv 74 weeks agoFavourite answer
v(t) = 12t^2-52t+56

Speed is the absolute value of velocity. The velocity, and speed, are 0 at t = 2 and

t =7/3.

Slowing down: (0, 2) ∪ (13/6, 7/3)

Speeding up: (2, 13/6) ∪ (7/3, 4)

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