Question

the graph below represents the speeds of karl and lyon as they run around a track. for how many minutes did lyon run at a greater speed than karl? ∘ 12 ∘ 17 ∘ 29 ∘ 20

Explanation:

Step1: Identify time intervals

Find when Lyon's speed > Kent's speed. From graph, intervals: (0,8), (8,12), (20,32)? Wait, recheck. Wait, Lyon's line (blue) and Kent's (black). Let's see key points: Lyon: (0,0), (8,16), (20,11), (32,2), (40,0). Kent: (0,0), (6,3), (12,12), (28,12), (40,0). So when is Lyon's speed > Kent's?

From t=0 to t=8: Lyon (up to 16) vs Kent (up to 3) → Lyon faster. Duration: 8 - 0 = 8.

From t=8 to t=12: Lyon at 16, Kent at 3 to 12. Wait, at t=8, Kent is at 3? Wait, Kent's points: (6,3), (12,12). So from t=6 to 12, Kent goes from 3 to 12. Lyon at t=8 is 16, t=12: Lyon is 11? Wait, Lyon's (20,11) – maybe typo, maybe (20,11) is Lyon? Wait, maybe the graph has Lyon (blue) with points (0,0), (8,16), (20,11), (32,2), (40,0). Kent (black): (0,0), (6,3), (12,12), (28,12), (40,0).

So when Lyon's speed > Kent's:

1. 0 to 8: Lyon (0→16) vs Kent (0→3) → Lyon faster. Time: 8 - 0 = 8.

2. 8 to 12: Lyon at 16 (t=8) to 11 (t=20? No, maybe Lyon's (20,11) is at t=20. Wait, maybe the x-axis is time (minutes) from 0 to 40. Let's list time when Lyon's speed (y) > Kent's (y):

- From t=0 to t=8: Lyon's y > Kent's (Kent at t=8: let's calculate Kent's speed at t=8. Kent goes from (6,3) to (12,12), so slope is (12-3)/(12-6)= 9/6=1.5. So at t=8, Kent's speed is 3 + 1.5*(8-6)= 3 + 3=6. Lyon at t=8 is 16. So 16>6: Lyon faster. Time: 8-0=8.

- From t=8 to t=12: Kent's speed at t=12 is 12. Lyon at t=12: let's see Lyon's slope from (8,16) to (20,11): (11-16)/(20-8)= -5/12 ≈ -0.4167. So at t=12, Lyon's speed is 16 + (-5/12)*(12-8)= 16 - 5/3 ≈ 14.333. Kent at t=12 is 12. So 14.333>12: Lyon faster. Time: 12-8=4.

- From t=12 to t=20: Kent's speed is 12 (constant from t=12 to t=28). Lyon's speed at t=12: ~14.333, at t=20:11. So 14.333>12 (t=12-20? Wait, 11 at t=20: 11<12? Wait, at t=20, Lyon's speed is 11, Kent's is 12. So at t=20, Lyon's speed is less. So when does Lyon's speed drop below 12? Let's solve 16 - (5/12)(t-8) = 12 → (5/12)(t-8)=4 → t-8= (4*12)/5= 9.6 → t=17.6? Wait, maybe the graph is simpler. The options are 12,17,29,20. Wait, maybe the correct intervals are:

Lyon's faster time: from t=0 to t=8 (8 min), t=8 to t=12 (4 min), and maybe another? Wait, no, maybe the graph has Lyon (blue) with points (0,0), (8,16), (20,11), (32,2), (40,0) and Kent (black) (0,0), (6,3), (12,12), (28,12), (40,0). So the time when Lyon's speed > Kent's:

- 0 to 8: 8 min.

- 8 to 12: 4 min (since at t=12, Kent is 12, Lyon is 16 - (5/12)*4 ≈ 14.33 >12).

- Then, from t=12 to t=28, Kent is 12. Lyon at t=20 is 11 (less than 12), so before t=20, when does Lyon's speed become 12? Let's calculate t when 16 - (5/12)(t-8)=12 → t-8= (4*12)/5=9.6 → t=17.6. So from t=8 to t=17.6, Lyon is faster? No, maybe the graph is designed with integer times. Wait, the options include 12,17,29,20. Let's check the difference in time when Lyon's speed is above Kent's.

Alternative approach: The area where Lyon's line is above Kent's. Let's list the time segments:

1. From 0 to 8: Lyon above Kent. Time: 8 - 0 = 8.

2. From 8 to 12: Lyon above Kent (since Kent at 12 is 12, Lyon at 12 is 16 - (5/12)*4 = 16 - 5/3 ≈14.33 >12). Time: 12 - 8 = 4.

3. From 12 to 28: Kent is at 12, Lyon is decreasing from ~14.33 to 11 (at t=20) then to 2 (at t=32). Wait, at t=20, Lyon is 11 <12, so from t=12 to t=20, Lyon is above Kent? No, 14.33 at t=12, 11 at t=20: so between t=12 and t=20, Lyon's speed is between 14.33 and 11, which is above 12? No, 11 is below 12. So when does Lyon's speed drop to 12? At t=12 + (14.33 -12)/(5/12) = 12 + (2.33)*(12/5) ≈12 + 5.6 =17.6. So from t=8 to t=17.6, Lyon is above Kent? Then total time: 8 (0-8) + (17.6 -8)= 17.6, close to 17. So the answer is 17.

Answer:

17