Two cars start out simultaneously from a point in the same direction, one of them going at a speed of 50 kmhr^-1 and the other at 40 kmhr-¹. In half an hour a third car starts out from the same point and overtakes the first car 1.5 hours after catching up with the second car. The speed of the third car is

Positions of the cars when third car starts is shown.

Let third car catches 2nd car in time T after it starts out, then

vT = 40 T + 20               (i)

Now third car catches 1st car after time T +1.5, so

V[T+1.5] = 50 [T+1.5]+25     (ii)

Put the value of T from equation (i) into equation (ii),

We get 1.5v2 -140v+3000 = 0

Put the value of T from equation (i) into equation (ii) , we get

= 3v2-280v +6000 = 0

$\mathrm{v}=\frac{280\pm \sqrt{{\left(280\right)}^2-72000}}{6}$

 $\Rightarrow \mathrm{v}=\frac{280\pm 80}{6}=60\mathrm{km}/\mathrm{h},\frac{100}{3}\mathrm{km}/\mathrm{h}$

Since v can’t be less than 50 km/h, we have 

V = 60 km/h