Question

motion • guided reading and study

1. the distance-versus-time graph above shows the motion of a jogger.

how far did the jogger run in 15 minutes? _______________

1. the distance-versus-time graph above also shows the motion of a jogger.

the line is divided into segments. the middle segment is horizontal.
what does that tell you about the jogger’s progress between minute 6
and minute 8?
_______________

Explanation:

Question 13

Step1: Analyze the graph

The distance - versus - time graph for the first jogger. We need to find the distance at \(t = 15\) minutes. By looking at the graph, we can see the point corresponding to \(t=15\) minutes on the x - axis and then find the corresponding y - value (distance) on the y - axis.
From the graph, when the time \(t = 15\) minutes, the distance \(d\) is read from the y - axis. The graph shows that at \(t = 15\) minutes, the distance is \(1300\) meters (by observing the grid and the position of the data point).

In a distance - versus - time graph, the slope of the line represents the speed of the object (\(speed=\frac{\text{change in distance}}{\text{change in time}}\)). A horizontal line segment means that the change in distance (\(\Delta d\)) is \(0\) (since the y - value, which is distance, does not change) while the change in time (\(\Delta t\)) is positive (from minute 6 to minute 8, \(\Delta t=8 - 6=2\) minutes). Since \(speed=\frac{\Delta d}{\Delta t}\) and \(\Delta d = 0\), the speed of the jogger is \(0\) during this time interval. This means the jogger was not moving (stationary) between minute 6 and minute 8.

Answer:

\(1300\) meters (or the value as per the accurate reading from the graph, if the grid is such that each major grid on y - axis is 200 meters and at \(t = 15\), the point is at 1300 meters approximately)

Question 14