Question

ninas run:
ryans run:
nina and ryan each ran at a constant speed for a 100-meter race. each runner’s distance for the same section of the race is displayed on the left. who had a head start, and how big was the head start?
had a head start of meters.
(graph for ninas run: points (4, 32), (6, 48), (8, 64), (10, 80) on a distance-time graph)
(table for ryans run: time (sec) - 4, 6, 8; distance (m) - 35, 47.5, 60)

Explanation:

Step1: Find Nina's distance at t=0

Nina's data points: (4,32), (6,48), (8,64), (10,80). The slope (speed) is $\frac{32}{4} = 8$ m/s. Using point - slope form $y - y_1=m(x - x_1)$, with (4,32) and $m = 8$, when $x = 0$, $y-32=8(0 - 4)\Rightarrow y=32-32 = 0$? Wait, no, wait. Wait, (4,32): if we consider the equation of the line. Let's take two points, say (4,32) and (8,64). The slope $m=\frac{64 - 32}{8 - 4}=\frac{32}{4}=8$. So the equation is $y=8x + b$. Plug in (4,32): $32=8\times4 + b\Rightarrow32 = 32 + b\Rightarrow b = 0$. Wait, but that would mean at $x = 0$, $y = 0$. Wait, no, maybe I made a mistake. Wait, (4,32): when $x = 4$, $y = 32$. So if we assume the line passes through the origin? Wait, no, let's check Ryan's data.

Step2: Find Ryan's distance at t = 0

Ryan's data: (4,35), (6,47.5), (8,60). The slope $m=\frac{47.5-35}{6 - 4}=\frac{12.5}{2}=6.25$ m/s. Equation: $y=6.25x + b$. Plug in (4,35): $35=6.25\times4 + b\Rightarrow35 = 25 + b\Rightarrow b = 10$. So at $x = 0$ (t = 0), Ryan's distance $y = 10$ meters.

Step3: Compare distances at t = 0

Nina's distance at t = 0: from her line, when x = 0, if we use the point (4,32) and slope 8, we saw $y = 0$. Wait, but maybe I misread Nina's graph. Wait, the graph for Nina: when x = 4 (time = 4 sec), y = 32 (distance = 32 m). So her speed is 32/4 = 8 m/s. So at t = 0, distance is 0. Ryan, at t = 0, distance is 10 m (from his equation $y = 6.25x+10$, when x = 0, y = 10). Wait, no, wait. Wait, maybe I messed up. Wait, let's check Ryan's data again. At t = 4, distance is 35. At t = 6, 47.5. The difference in time is 2 sec, difference in distance is 12.5 m. So speed is 12.5/2 = 6.25 m/s. So in 4 sec, he runs 6.25*4 = 25 m. But he has 35 m at t = 4, so the head start is 35 - 25 = 10 m. For Nina, at t = 4, she has 32 m. Her speed is 32/4 = 8 m/s, so in 4 sec, she runs 32 m, so head start is 0 (since 32-8*4 = 0). So Ryan has a head start? Wait, no, wait. Wait, maybe I got it backwards. Wait, no, let's think again. A head start means at t = 0, the distance from the start line. So if at t = 0, one has a non - zero distance, that's the head start.

Wait, Nina's graph: when t = 0, what's her distance? Let's see, the first point is at t = 4, d = 32. So if we extrapolate backwards, since her speed is 8 m/s, in 4 seconds, she runs 32 m, so at t = 0, she is at 0 m. Ryan's data: at t = 4, d = 35. His speed is 6.25 m/s, so in 4 seconds, he runs 6.25*4 = 25 m. So the distance he has at t = 0 is 35 - 25 = 10 m? Wait, no, that's not right. Wait, the equation of the line for Ryan: $y=mx + b$. We found m = 6.25, and when x = 4, y = 35. So $35=6.25*4 + b\Rightarrow35 = 25 + b\Rightarrow b = 10$. So the y - intercept (distance at t = 0) is 10. For Nina, let's find her y - intercept. Take two points (4,32) and (8,64). The slope is (64 - 32)/(8 - 4)=8. So equation is $y = 8x + b$. Plug in (4,32): $32=8*4 + b\Rightarrow32 = 32 + b\Rightarrow b = 0$. So at t = 0, Nina's distance is 0, Ryan's is 10? Wait, no, that can't be. Wait, maybe I mixed up who has the head start. Wait, no, wait. Wait, when t = 0, Ryan has 10 m, Nina has 0 m? No, that would mean Ryan has a head start. But wait, let's check at t = 4. Nina has 32 m, Ryan has 35 m. At t = 6, Nina has 48 m, Ryan has 47.5 m. At t = 8, Nina has 64 m, Ryan has 60 m. So Nina is catching up. So at t = 0, Ryan was at 10 m, Nina at 0 m? Wait, no, maybe my calculation for Ryan's y - intercept is wrong. Wait, let's recalculate Ryan's slope. From (4,35) to (8,60): the change in x is 4, change in y is 25. So slope is 25/4 = 6.25. Then equation: $y=6.25x + b$. At x = 4, y = 35: $35=6.25*4 + b\Rightarrow35 = 25 + b\Rightarrow b = 10$. So at x = 0, y = 10. For Nina, at x = 0, y = 0. So Ryan had a head start of 10 meters? Wait, no, wait, maybe I made a mistake. Wait, let's check Nina's graph again. The first point is at (4,32), so when time is 4 seconds, distance is 32 meters. So her speed is 32/4 = 8 m/s. So if we go back to time 0, she would have run 0 meters (since 8 m/s * 0 s = 0 m). Ryan, at time 4 seconds, has run 35 meters. His speed is 35/4 = 8.75? Wait, no, earlier calculation was wrong. Wait, (4,35), (6,47.5): 47.5 - 35 = 12.5, 6 - 4 = 2, 12.5/2 = 6.25. (6,47.5), (8,60): 60 - 47.5 = 12.5, 8 - 6 = 2, 12.5/2 = 6.25. So speed is 6.25 m/s. So in 4 seconds, he runs 6.25*4 = 25 m. But he has 35 m at t = 4, so the head start is 35 - 25 = 10 m. For Nina, in 4 seconds, she runs 32 m, so no head start (32 - 8*4 = 0). So Ryan had a head start of 10 meters? Wait, but wait, the problem says "who had a head start". Wait, maybe I got it backwards. Wait, let's check the y - intercepts. The y - intercept of the distance - time graph is the initial distance (head start). Nina's line passes through (4,32) and has slope 8, so equation y = 8x, y - intercept 0. Ryan's line has equation y = 6.25x+10, y - intercept 10. So Ryan had a head start of 10 meters? Wait, no, wait, maybe I made a mistake in Nina's equation. Wait, (4,32): if x = 0, y = 0. (8,64): 8*8 = 64, yes. (10,80): 8*10 = 80, yes. So Nina's equation is y = 8x, so at t = 0, y = 0. Ryan's equation: let's take (4,35): 6.25*4+10 = 25 + 10 = 35, correct. (6,47.5): 6.25*6+10 = 37.5+10 = 47.5, correct. (8,60): 6.25*8+10 = 50+10 = 60, correct. So Ryan's initial distance (at t = 0) is 10 m, Nina's is 0 m. So Ryan had a head start of 10 meters? Wait, but wait, when t = 4, Nina has 32 m, Ryan has 35 m. So Ryan is ahead at t = 4, but Nina is faster (8 m/s vs 6.25 m/s). So the head start is the initial distance, so Ryan had a head start of 10 meters. Wait, no, wait, maybe I messed up. Wait, the problem is about who had a head start. A head start means that at t = 0, one runner is already ahead (has a non - zero distance). So Ryan's distance at t = 0 is 10 m, Nina's is 0 m. So Ryan had a head start of 10 meters? Wait, but let's check again. Wait, maybe Nina's graph: is the first point at (4,32) or is there a point at t = 0? No, the graph starts at t = 4. Wait, maybe the x - axis starts at t = 0, but the first point is at t = 4. Wait, no, the x - axis is time (sec), with t = 0 at the origin. So the points for Nina are at t = 4,6,8,10. So to find the head start, we need to find the distance at t = 0 for both. For Nina, since her speed is constant, we can find the equation of the line. As we did, y = 8x, so at t = 0, y = 0. For Ryan, using his table, we found the equation y = 6.25x+10, so at t = 0, y = 10. So Ryan had a head start of 10 meters. Wait, but wait, maybe I made a mistake in Ryan's equation. Let's recalculate Ryan's slope. From t = 4 to t = 8, time increases by 4, distance increases by 60 - 35 = 25. So slope is 25/4 = 6.25. Correct. Then equation: y = 6.25x + b. At t = 4, y = 35: 35 = 6.25*4 + b => 35 = 25 + b => b = 10. Correct. So Ryan's initial distance is 10 m, Nina's is 0 m. So Ryan had a head start of 10 meters.

Answer:

Ryan had a head start of 10 meters.