Question

Two cars start together in same direction from the same place the first goes with uniform speed of 10km/h . The second goes at a speed of 8km/h n the first hour and increases the speed by by 1/2km/h in each succeeding hour . After how many hours will the second car overtake the first ? 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

$$\begin{array}{cc}& 10t=\frac{t(t+31)}{4}\Rightarrow 40t=t^2+31t\Rightarrow t^2-9t=0\\ & \Rightarrow t(t-9)=0\Rightarrow t=9[\because t\equiv 0]\end{array}$$ Thus, the second car will overtake the first car in 9 hours.