Yuto left his house at 10 a.m. to go for a bike ride. By the time Yuto’s sister Riko left their house, Yuto was already 5.25 miles along the path they both took. If Yuto’s average speed was 0.25 miles per minute and Riko’s average speed was 0.35 miles per minute, over what time period in minutes, t, starting from when Riko left the house, will Riko be behind her brother?

Question

Riko will be behind Yuto when 0 ≤ t <
minutes.

Answer 1

To determine the time period in minutes, \( t \), during which Riko will be behind Yuto after she leaves the house, we need to compare their positions as functions of time.

1. Calculate Yuto's Position: Yuto started biking at 10 a.m. and is already 5.25 miles ahead when Riko leaves. His speed is 0.25 miles per minute. If \( t \) is the time in minutes after Riko leaves, Yuto's position at that time can be expressed as:

   \[ \text{Yuto's position} = 5.25 + 0.25t \]

2. Calculate Riko's Position: Riko's speed is 0.35 miles per minute. Since she leaves the house at the moment \( t = 0 \), her position can be expressed as:

   \[ \text{Riko's position} = 0.35t \]

3. Set Up the Inequality: Riko will be behind Yuto when:

   \[ 0.35t < 5.25 + 0.25t \]

   Rearranging this inequality gives:

   \[ 0.35t - 0.25t < 5.25 \]

   \[ 0.10t < 5.25 \]

4. Solve for \( t \):

   Dividing both sides by 0.10:

   \[ t < \frac{5.25}{0.10} \]

   \[ t < 52.5 \]

Thus, Riko will be behind Yuto when:

\[ 0 \leq t < 52.5 \text{ minutes} \]

Final answer:

\[ \text{Riko will be behind Yuto when } 0 \leq t < 52.5. \]