Question

1. reason abstractly explain why a student who runs \\(\frac{3}{4}\\) mile in 6 minutes is faster than a student who runs \\(\frac{1}{2}\\) mile in 5 minutes.

Explanation:

Step1: Calculate speed of first student

Speed is distance divided by time. For the first student, distance is $\frac{3}{4}$ mile and time is 6 minutes. So speed $v_1 = \frac{\frac{3}{4}}{6}=\frac{3}{4}\times\frac{1}{6}=\frac{3}{24}=\frac{1}{8}$ miles per minute.

Step2: Calculate speed of second student

For the second student, distance is $\frac{1}{2}$ mile and time is 5 minutes. So speed $v_2=\frac{\frac{1}{2}}{5}=\frac{1}{2}\times\frac{1}{5}=\frac{1}{10}$ miles per minute.

Step3: Compare the two speeds

Compare $\frac{1}{8}$ and $\frac{1}{10}$. Since the numerators are the same, the fraction with the smaller denominator is larger. So $\frac{1}{8}>\frac{1}{10}$, meaning the first student's speed is higher.

Answer:

The first student (who runs $\frac{3}{4}$ mile in 6 minutes) has a speed of $\frac{1}{8}$ miles per minute, and the second student (who runs $\frac{1}{2}$ mile in 5 minutes) has a speed of $\frac{1}{10}$ miles per minute. Since $\frac{1}{8}>\frac{1}{10}$, the first student is faster.