kinematics question

Question

A student starts from his house with a speed of 12 kmh^-1 and reaches the school 3 min late. Next day he increases speed by 1 kmh^-1 and reaches the school 3 min earlier. Find the distance between his house and office.

Answer 1

Let distance b/w house and school = d km

Let time required to reach school be T hour

So, according to question

Increase of Speed by 1 km/h

Substracting above, we get

\(\frac{d}{12}-\frac{d}{13}=\frac{3}{60}+\frac{3}{60}\)

\(\frac{(13-12)d}{12\times13}=\frac{6}{60}\)

\(\frac{d}{12\times 13}=\frac{1}{10}\)

\(d=\frac{12\times 13}{10}=15.6\,km\)