Question

nina and ryan each ran at a constant speed for a 100 - meter race. each runners 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. ninas run: (4, 32), (6, 48), (8, 64), (10, 80) ryans run: time (sec) distance (m) 4 35 6 47.5 8 60

Explanation:

Step1: Find Nina's distance at time t = 0

First, find the equation of the line for Nina's run. The slope $m$ using two - points $(x_1,y_1)=(4,32)$ and $(x_2,y_2)=(6,48)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{48 - 32}{6 - 4}=\frac{16}{2}=8$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(4,32)$ and $m = 8$, we have $y-32=8(x - 4)$. Simplifying gives $y=8x$. When $x = 0$, $y = 0$.

Step2: Find Ryan's distance at time t = 0

Find the slope of Ryan's line using two points $(x_1,y_1)=(4,35)$ and $(x_2,y_2)=(6,47.5)$. The slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{47.5-35}{6 - 4}=\frac{12.5}{2}=6.25$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(4,35)$ and $m = 6.25$, we get $y-35=6.25(x - 4)$. Expanding: $y-35=6.25x-25$, so $y=6.25x + 10$. When $x = 0$, $y = 10$.

Answer:

Ryan; 10