SSC GD Constable 6 Feb 2025 Shift 2 Paper

Relative speed $($ opposite direction $)={\mathit{S}}_1+{\mathit{S}}_2$
Distance $=$ Speed × Time
$1\mathit{k}\mathit{m}∕\mathit{h}=5∕18\mathit{m}∕\mathit{s}$
When two trains cross each other, the total distance covered is the sum of their lengths.
Calculation:
Relative speed $=50\mathit{k}\mathit{m}∕\mathit{h}+110\mathit{k}\mathit{m}∕\mathit{h}=160\mathit{k}\mathit{m}∕\mathit{h}$
Convert relative speed to $\mathit{m}∕\mathit{s}$ :
$\Rightarrow 160\mathit{k}\mathit{m}∕\mathit{h}=160\times \frac{5}{18}\mathit{m}∕\mathit{s}$
⇒ Relative speed $=\frac{800}{18}\mathit{m}∕\mathit{s}=\frac{400}{9}\mathit{m}∕\mathit{s}$
Let the length of train 2 be ${\mathit{L}}_2$.
Total distance covered $={\mathit{L}}_1+{\mathit{L}}_2=500+{\mathit{L}}_2$ meters
Using Distance $=$ Speed × Time:
$\Rightarrow 500+{\mathit{L}}_2=\frac{400}{9}\times 12$
$\Rightarrow 500+{\mathit{L}}_2=\frac{400\times 4}{3}$
$\Rightarrow 500+{\mathit{L}}_2=\frac{1600}{3}$
$\Rightarrow 500+{\mathit{L}}_2=533.33$
$\Rightarrow {\mathit{L}}_2=533.33-500$
$\Rightarrow {\mathit{L}}_2=33.33$