Question

Two cars start together in the same direction from the same place. The first goes with uniform speed of $10km/h$. The second goes at a speed of $8km/h$ in the first hour and increases the speed by $\frac{1}{2}km$ in each succeeding hour. After how many hours will the second car overtake the first car if both cars go non-stop?

Answer 1

Suppose the second car overtakes the first car after $t$ hours.

Then, the two cars travel the same distance in $t$ hours.

Distance travelled by the first car in $t$ hours $=10tkm$.

Distance travelled by the second car in $t$ hours.
$=$ Sum of $t$ terms of an A.P. with first term $8$ and common difference $1/2$.
$=\frac{t}{2}\{2\times 8+(t-1)\times \frac{1}{2}\}=\frac{t(t+31)}{4}$
When the second car overtakes the first car, we have
$10t=\frac{t(t+31)}{4}⟹40t=t^2+31t⟹t^2-9t=0⟹t(t-9)=0⟹t=9$ $[\because t\ne 0]$
Thus, the second car will overtake the first car in $9$ hours.