Time and Distance - Q1 - To and Fro Motion in a Straight Line

Q. Ram and Gopi run towards each other starting from K and L respectively with respective speeds of 2 kmph and 3 kmph. After meeting each other, to reach L, if Ram takes 7 hours less than the square of the magnitude of the time (in hours) taken by Gopi to reach K, find distance b/w K and L.

A.

Let M be the Meeting Point of Ram and Gopi.
Let KM = x and ML = y.

Let time taken to meet be 't' hours.
Therefore,
Distance traveled by Ram in time t = KM = x = 2t km
Distance traveled by Gopi in time t = ML = y = 3t km

Hence,
$$\frac{x}{y}=\frac{2}{3}$$
Hence, Let x = 2k and y = 3k.
After meeting at time t,
let t_R be the time taken by Ram to reach L, and let t_G be the time taken by Gopi to reach K

$$t_R=\frac{\text{Distance traveled by Ram}}{\text{Speed of Ram}}=\frac{y}{2}$$

$$t_G=\frac{\text{Distance traveled by Gopi}}{\text{Speed of Gopi}}=\frac{x}{3}$$

From "to reach L, if Ram takes 7 hours less than the square of the magnitude of the time (in hours) taken by Gopi to reach K",

$$t_R={\left(t_G\right)}^2-7$$ $$\begin{array}{l}\to \frac{y}{2}={\left(\frac{x}{3}\right)}^2-7\\ \to 2x^2-9y-126=0\end{array}$$

Substituting for x and y as 2k and 3k,
$$\begin{array}{l}\to {\left(2k\right)}^2-9\left(3k\right)-126=0\\ \to 4k^2-27k-126=0\end{array}$$

Solving for k,
$$\begin{array}{l}k=\frac{27\pm \sqrt{{27}^2-\left[\left(4\right)\times \left(8\right)\times \left(-126\right)\right]}}{2\times 8}\\ \\ \to k=\frac{27\pm \sqrt{729+4032}}{16}=\frac{27\pm 69}{16}\\ \\ \to k=6\text{ OR }-\left(\raisebox{1ex}{$21$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right)\end{array}$$

Therefore,
$$\begin{array}{l}x=2k=12\\ y=3k=18\end{array}$$

Distance between K & L,
$$\text{KL}$$

Therefore the distance between K & L is 30 kms.