Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km h^-1. The second car goes at a speed of 8 km h^-1 in the first hour and thereafter increasing the speed by 0.5 km h^-1 each succeeding hour. After how many hours will the two cars meet ?

Let the two cars meet after n hours.

Distance covered by first car = 10n km

Since, second car goes at a speed of 8 km h^-1 in the first hour and thereafter increasing the speed by 0.5 km h^-1 each succeeding hour

Hence, its an A.P. with n terms (as there are n hours).

Distance covered by second car = $\dfrac{n}{2}[2a + (n - 1)d] = \dfrac{n}{2}[2 \times 8 + (n - 1) \times 0.5]$

Since, cars meet after n hours.

$\therefore 10n = \dfrac{n}{2}[2 \times 8 + (n - 1) \times 0.5] \\[1em] \Rightarrow \dfrac{n}{2}[16 + 0.5n - 0.5] = 10n \\[1em] \Rightarrow 15.5 + 0.5n = 20 \\[1em] \Rightarrow 0.5n = 4.5 \\[1em] \Rightarrow n = 9.$

Hence, car will meet after 9 hours.