Question

choose the best answers using the distance - time graph. three runners are in a race. 19. choose the best answer. which table represents the movement of the runners shown by line c? 20. choose the best answer. which runner moved 1300 meters in 9 minutes? runner c runner b runner a

Explanation:

Step1: Analyze the speed of Line C

The speed \(v=\frac{d}{t}\). For the first table: \(v_1 = \frac{88}{2}=44\space m/s\), \(v_2=\frac{176}{4} = 44\space m/s\), \(v_3=\frac{264}{6}=44\space m/s\), \(v_4=\frac{352}{8}=44\space m/s\).
For the second table: \(v_1=\frac{200}{2} = 100\space m/s\), \(v_2=\frac{400}{4}=100\space m/s\), \(v_3=\frac{600}{6}=100\space m/s\), \(v_4=\frac{800}{8}=100\space m/s\).
For the third table: \(v_1=\frac{295}{2}=147.5\space m/s\), \(v_2=\frac{590}{4}=147.5\space m/s\), \(v_3=\frac{895}{6}\approx149.17\space m/s\), \(v_4=\frac{1180}{8}=147.5\space m/s\).
From the graph, Line C has a smaller slope (less steep), indicating a slower speed. The first table has a speed of \(44\space m/s\), which is consistent with the slower - moving Line C.

Step2: Analyze the second question

From the graph, at \(t = 9\) minutes (\(t=9\times60 = 540\space s\)), Runner A has a distance of approximately \(1600\space m\), Runner B has a distance of approximately \(1000\space m\), and Runner C has a distance of approximately \(500\space m\). Wait, there is a mistake. Let's re - check.
Wait, the second question: we can also use the slope concept. The slope of the distance - time graph is speed \(v=\frac{\Delta d}{\Delta t}\).
For Runner A: assume two points \((0,0)\) and \((10,1600)\) (approximate from the graph), \(v_A=\frac{1600}{10\times60}=\frac{8}{3}\space m/s\approx2.67\space m/s\). In \(t = 9\times60=540\space s\), \(d_A=v_A\times t=\frac{8}{3}\times540 = 1440\space m\) (approximate).
For Runner B: assume two points \((0,0)\) and \((10,1100)\) (approximate from the graph), \(v_B=\frac{1100}{10\times60}=\frac{11}{6}\space m/s\approx1.83\space m/s\). In \(t = 9\times60 = 540\space s\), \(d_B=v_B\times t=\frac{11}{6}\times540=990\space m\) (approximate).
For Runner C: assume two points \((0,0)\) and \((10,500)\) (approximate from the graph), \(v_C=\frac{500}{10\times60}=\frac{5}{6}\space m/s\approx0.83\space m/s\). In \(t = 9\times60=540\space s\), \(d_C=v_C\times t=\frac{5}{6}\times540 = 450\space m\) (approximate). Wait, no. Wait, maybe the units on the time - axis of the graph: if we consider the scale.
Wait, another approach: look at the vertical axis (distance) and horizontal axis (time). At \(t = 9\) (on the time - axis of the graph, assume the scale is 1 unit on the time - axis is 1 minute). For Runner A, at \(t = 9\), the distance is approximately \(1300\space m\) (by visual inspection of the graph).

Answer:

1. The first table (time (s): 2,4,6,8; distance (m):88,176,264,352) represents Line C.
2. Runner A.