Question: A train is timmed to run from Howrah to Delhi at an average speed of $u$ kilometers per hour. Due to some engine trouble the starting of the train was delayed by $h$ hours. At Delhi the speed is raised to $v$ kilometers per hour in order that the train may reach in time. If the train runs with this increased speed up to a station at a distance $s$ kilometers from Delhi, then show that $s(v-u)=uvh$.
My approach: Let $d$ be the total distance between Howrah(H) to Delhi(D). Total time $t=\dfrac{d}{u}$. Let at point $P$, it raised velocity $v$ km/h. Then time from H to P is $t_1=\dfrac{d-s}{u}$ and time from P to D is $t_2=\dfrac{s}{v+u}$.
I don't know whether my approach is true or not. I can I get the require result?
I assume that the goal of the train , is to reach the station $s$ kilometers from Delhi . Another important point to note is that the speed is raised to $v$ at Delhi .
Let $d$ be the total distance to Delhi(D) from Howrah(H) . Let the total time $=t$ . Then , we have :- $$ t = \frac{d+s}{u} = \frac{d}{u} + \frac{s}{u} \tag{1} $$ $$ t - h = \frac{d}{u} + \frac{s}{v} \tag{2} $$ From $(1),(2),$ we get :- $$ h = \frac{s}{u}-\frac{s}{v} \implies \boxed{uvh=s(v-u)}$$