Question

in which segment does the object have the highest average speed? the distance - time graph of an object is shown. distance (km) time (hours) clear all segment de segment cd segment bc segment ab

Explanation:

Step1: Recall speed - distance - time formula

Average speed $v=\frac{\Delta d}{\Delta t}$, where $\Delta d$ is the change in distance and $\Delta t$ is the change in time.

Step2: Analyze segment AB

Let the starting - point of segment AB be $(t_1 = 0,h_1 = 0)$ and the ending - point be $(t_2=6,h_2 = 6)$. Then $\Delta d_{AB}=6 - 0=6$ km and $\Delta t_{AB}=6 - 0 = 6$ h. So, $v_{AB}=\frac{6}{6}=1$ km/h.

Step3: Analyze segment BC

The starting - point of segment BC is $(t_1 = 6,h_1 = 6)$ and the ending - point is $(t_2 = 9,h_2 = 9)$. Then $\Delta d_{BC}=9 - 6=3$ km and $\Delta t_{BC}=9 - 6 = 3$ h. So, $v_{BC}=\frac{3}{3}=1$ km/h.

Step4: Analyze segment CD

The starting - point of segment CD is $(t_1 = 9,h_1 = 9)$ and the ending - point is $(t_2 = 12,h_2 = 12)$. Then $\Delta d_{CD}=12 - 9=3$ km and $\Delta t_{CD}=12 - 9 = 3$ h. So, $v_{CD}=\frac{3}{3}=1$ km/h.

Step5: Analyze segment DE

The starting - point of segment DE is $(t_1 = 12,h_1 = 12)$ and the ending - point is $(t_2 = 16,h_2 = 16)$. Then $\Delta d_{DE}=16 - 12=4$ km and $\Delta t_{DE}=16 - 12 = 4$ h. So, $v_{DE}=\frac{4}{4}=1$ km/h.

Since the average speed in all segments is 1 km/h, all segments have the same average speed. But if we consider the non - zero slope segments only (assuming no other unmentioned conditions), we note that the slope of the distance - time graph represents speed. And visually, all the non - horizontal segments have the same slope.

Answer:

All segments have the same average speed. If we must choose one, any of Segment AB, Segment BC, Segment CD, Segment DE would be correct as they all have an average speed of 1 km/h.