Relative Motion of Train and Boy
Let $v$ be the speed of the train and $u = 10$ km/h be the speed of the boy. Let $L_c$ be the length of the coach and $s=1$ m be the step length.
(a) Speed of the train
The boy counts steps while passing the coach. This implies he covers the relative distance equal to the length of the coach $L_c$.
Case 1: Moving in same direction
The boy is faster than the train (since he “passes” it). Relative speed is $u – v$.
Distance covered by boy: $d_1 = n_1 s = 30 \times 1 = 30$ m.
Time taken: $t_1 = d_1 / u$.
Length of coach: $L_c = (u – v) t_1 = (u – v) \frac{n_1 s}{u}$.
Case 2: Moving in opposite direction
Relative speed is $u + v$.
Distance covered by boy: $d_2 = n_2 s = 20 \times 1 = 20$ m.
Time taken: $t_2 = d_2 / u$.
Length of coach: $L_c = (u + v) t_2 = (u + v) \frac{n_2 s}{u}$.
Equating $L_c$ from both cases:
$$ (u – v) n_1 = (u + v) n_2 $$ $$ u(n_1 – n_2) = v(n_1 + n_2) $$ $$ v = u \left( \frac{n_1 – n_2}{n_1 + n_2} \right) $$Substituting values ($u=10, n_1=30, n_2=20$):
$$ v = 10 \left( \frac{30 – 20}{30 + 20} \right) = 10 \left( \frac{10}{50} \right) = 2 \text{ km/h} $$(b) Length of the coach
Using the equation from Case 2:
Answers:
(a) 2 km/h
(b) 24 m