# Can you help me interpret the motion of the following position-time graph?

## I want to interpret the motion of each slope in this p-t graph. For example, considering that the position is in (m) [N], I would say that the first slope represents an object moving [N] at a constant velocity. But, how would I interpret slopes that go are situated below the x-axis?

Oct 19, 2016

As we see in the given graph, there are eight time segments
1. $t = 0$ to $t = 2$.
We see that slope of the line is given by $\frac{2 m}{2 s} = 1 m {s}^{-} 1$.
Implies that the object is moving in the positive direction with a constant velocity of $1 m {s}^{-} 1$.
2. $t = 2$ to $t = 6$.
We see that slope of the line is given by $\frac{2 m}{4 s} = 0.5 m {s}^{-} 1$.
Implies that the object is moving in the positive direction with a constant velocity of $0.5 m {s}^{-} 1$.
3. $t = 6$ to $t = 8$.
We see that slope of the line is given by $\frac{0 m}{2 s} = 0 m {s}^{-} 1$.
Implies that the object is stationary
4. $t = 8$ to $t = 12$.
We see that slope of the line is given by $\frac{- 8 m}{4 s} = - 2 m {s}^{-} 1$.
Implies that the object is moving in the negative direction with a constant velocity of $2 m {s}^{-} 1$.
5. $t = 12$ to $t = 14$.
We see that slope of the line is same as in step 3.
The object is stationary.
6. $t = 14$ to $t = 16$.
We see that slope of the line is given by $\frac{2 m}{2 s} = 1 m {s}^{-} 1$.
Implies that the object is moving in the positive direction with a constant velocity of $1 m {s}^{-} 1$.
7. $t = 16$ to $t = 18$.
We see that slope of the line is given by $\frac{4 m}{2 s} = 2 m {s}^{-} 1$.
Implies that the object is moving in the positive direction with a constant velocity of $2 m {s}^{-} 1$.
8. $t = 18$ to $t = 20$.
We see that slope of the line is same as in steps 3 and 5.
The object is stationary.