What is the speed of a train if it overtakes two persons who are walking in the same direction at the rate of $a$ m/s and $(a + 1)$ m/s and passes them completely in $b$ seconds and $(b + 1)$ seconds respectively? (C.D.S., 2004)

### Concept & Algebraic Formulation When a train overtakes objects moving in the same direction, the distance covered equals the train's length, and the effective speed is the difference between their speeds. $$ L = (V - u) \times t $$ ### Step-by-Step Solution * Let the speed of the train be $v$ m/s and its length be $L$ meters. * The train passes the first person moving at $a$ m/s in $b$ seconds. So, $L = (v - a) \times b$. * The train passes the second person moving at $(a + 1)$ m/s in $(b + 1)$ seconds. So, $L = (v - (a + 1)) \times (b + 1)$. * Equating the lengths: $(v - a)b = (v - a - 1)(b + 1)$. * Expanding both sides: $vb - ab = vb + v - ab - a - b - 1$. * Canceling common terms ($vb$ and $-ab$): $0 = v - a - b - 1$. * Solving for $v$: $v = a + b + 1$. ### Exam Strategy & Shortcut For a general pattern where speeds increase by 1 unit and times increase by 1 unit, the train's speed is always the sum of the initial speed, initial time, and the increment. ### Common Pitfall Making algebraic sign errors when expanding the terms $(v - (a + 1))(b + 1)$. Always use brackets carefully. ### Final Answer Therefore, the correct answer is **$(a + b + 1)$ m/s**.