Question

how does a distance vs. time graph represent an object at rest? (1 point) with an upward-sloping line segment with a downward-sloping line segment with a vertical line segment with a horizontal line segment

Explanation:

In a distance - time graph, the slope of the line segment represents the speed of the object. The formula for speed is \(v=\frac{\Delta d}{\Delta t}\), where \(v\) is speed, \(\Delta d\) is the change in distance, and \(\Delta t\) is the change in time. For an object at rest, the distance \(d\) does not change with time \(t\), so \(\Delta d = 0\).

• An upward - sloping line segment: Here, \(\Delta d>0\) as time \(t\) increases. Using the speed formula \(v = \frac{\Delta d}{\Delta t}\), since \(\Delta d>0\) and \(\Delta t>0\), \(v>0\) (the object is moving forward).
• A downward - sloping line segment: Here, \(\Delta d<0\) (distance is decreasing) as time \(t\) increases. Using the speed formula \(v=\frac{\Delta d}{\Delta t}\), since \(\Delta d < 0\) and \(\Delta t>0\), \(v<0\) (the object is moving backward).
• A vertical line segment: Here, \(\Delta t = 0\) (time is not passing) while \(\Delta d

eq0\). The formula \(v=\frac{\Delta d}{\Delta t}\) would result in an undefined value (division by zero), which is not physically meaningful for the motion of an object.

• A horizontal line segment: Here, \(\Delta d=0\) as time \(t\) increases. Using the speed formula \(v=\frac{\Delta d}{\Delta t}\), since \(\Delta d = 0\) and \(\Delta t>0\), \(v = 0\). This means the object is at rest.

Answer:

with a horizontal line segment