Question

Two cars start together in the same direction from the same place. The first car goes at uniform speed of $10$ kmph. The second car goes at a speed of $8$ kmph in the first hour and thereafter increasing the speed by $0.5$ kmph each succeeding hour. After how many hours will the two cars meet?

Answer 1

Que : two car start together in same direction

first uniform speed = 10 km/h $d=10\times time$

Second car speed = 8 lm /h $=10\times n$

$\to$ distance covered by first = distance covered by second

$\therefore 10\times n=8+(8+\frac{1}{2})+(8+\frac{2}{2})+(8+\frac{3}{2})+----+(8+(\frac{n-1}{2}\left)\right)$

$\therefore 10\times n=[8+8+8+8----]+[\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+---+\frac{n-1}{2}]$

$=8n+\frac{1}{2}[1+2+3+---(n-1\left)\right]$

$\therefore 10n=8n+\frac{1}{2}\frac{n(n-1)}{2}$

$\therefore 10n-8n=\frac{1}{4}n(n-1)$

$\therefore 2n=\frac{1}{4}n(n-1)$

$\therefore 8=n-1\Rightarrow n=9$