Question

Luiz and Bianca are brother and sister and go to the same school. Luiz walks to school using a footpath, a distance of 900 m, and he walks at 1.5 m/s. Bianca cycles to school along the roads, a distance of 1500 m, and she cycles at 5 m/s. They both plan on arriving at school at 8:55 AM. Who leaves home first and by how much?

Answer 1

Answer:

Luiz left home first, 10 minutes before arriving time.

Explanation:

We assume that both Luiz and Bianca left the homes and travelled at constant speed. The purpose of this problem is determining who left home first, which is related to time. From Kinematics, both can be represented by the following expression:

$\Delta t = \frac{s}{v}$ (Eq. 1)

Where:

$s$ - Travelled distance, measured in meters.

$v$ - Speed, measured in meters per second.

$\Delta t$ - Time, measured in seconds.

Then, times taken by each person is now calculated:

Luiz ($s = 900\,m$, $v = 1.5\,\frac{m}{s}$)

$\Delta t_{L} = \frac{900\,m}{1.5\,\frac{m}{s} }\times \frac{1}{60}\,\frac{min}{s}$

$\Delta t_{L} = 10\,min$

Bianca ($s = 1500\,m$, $v = 5\,\frac{m}{s}$)

$\Delta t_{B} = \frac{1500\,m}{5\,\frac{m}{s} }\times \frac{1}{60}\,\frac{min}{s}$

$\Delta t_{B} = 5\,min$

If both arrived at school at 8:55 AM, then Luiz left home at 8:45 AM. Luiz left home first, 10 minutes before arriving time.