Question

Two cars start out simultaneously from a point in the same direction, one of them going at a speed of $50kmh{r}^{-1}$ and the other at $40kmh{r}^{-1}$. 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

• A. $55kmh{r}^{-1}$
• B. $60kmh{r}^{-1}$
• C. $72kmh{r}^{-1}$
• D. $90kmh{r}^{-1}$

Answer 1

Answer: (B)

Positions of the cars when third car starts is shown.


Let third car catches $2$nd car in time $T$ after it starts out, then
$vT=40T+20(i)$
Now third car catches $1$st 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 $1.5v^2-140v+3000=0$
Put the value of $T$ from equation $(i)$ into equation $(ii)$, we get
$\Rightarrow 2v^2-280v+6000=0$
$\Rightarrow v=\frac{280\pm \sqrt{(280{)}^2-72000}}{6}$
$\Rightarrow v=\frac{280\pm 80}{6}$
$=60km/h,\frac{100}{3}km/h$
Since $v$ can’t be less than $50km/h$, we have
$v=60km/h$