Answer
(a) The particle is speeding up on the intervals $(1,2)$ and $(3,4)$
The particle is slowing down on the intervals $(0,1)$ and $(2,3)$
(b) The particle is speeding up on the intervals $(1,2)$ and $(3,4)$
The particle is slowing down on the intervals $(0,1)$ and $(2,3)$
Work Step by Step
$v(t) = \frac{ds}{dt}$
Velocity is the slope of the position versus time graph. When the slope is getting steeper, then the particle is speeding up. When the slope is getting less steep, then the particle is slowing down.
(a) The slope is getting steeper on the intervals $(1,2)$ and $(3,4)$
Therefore, the particle is speeding up on the intervals $(1,2)$ and $(3,4)$
The slope is getting less steep on the intervals $(0,1)$ and $(2,3)$
Therefore, the particle is slowing down on the intervals $(0,1)$ and $(2,3)$
(b) The slope is getting steeper on the intervals $(1,2)$ and $(3,4)$
Therefore, the particle is speeding up on the intervals $(1,2)$ and $(3,4)$
The slope is getting less steep on the intervals $(0,1)$ and $(2,3)$
Therefore, the particle is slowing down on the intervals $(0,1)$ and $(2,3)$
