Q.A metro train runs from station A to B to C. It takes 4 minutes in travelling from station A to station B. The train halts at station B for 20 s. Then it starts from station B and reaches station C in the next 3 minutes. At the start, the train accelerates for 10 s to reach the constant speed of 72 km/hr. The train moving at the constant speed is brought to rest in 10 s at the next station.

[!TLDR] Break each leg into accelerate (10s) + constant speed + decelerate (10s) phases; distance in each phase is the area un …

Constant running speed $=72$ km/hr $=20$ m/s. A to B (total 4 min $=240$ s): accelerates from 0 to 20 m/s in 10 s — distance (average speed × time) $=\tfrac12(0+20)\times10=100$ m; runs at constant 20 m/s for the remaining $240-10-10=220$ s — distance $=220\times20=4400$ m; decelerates from 20 m/s to 0 in 10 s — distance $=\tfrac12(20+0)\times10=100$ m. Total AB $=100+4400+100=4600$ m $=4.6$ km. B to C (total 3 min $=180$ s, same 10 s accelerate/decelerate phases): constant-speed duration $=180-10-10=160$ s — …

Model each leg of the journey as a trapezoidal v-t profile (accelerate for 10 s, cruise at constant 72 km/hr, decelerate for 10 s); compute the distance in each phase as …

Including the 20 s halt time at station B as part of either leg's travel time (it does not contribute to either AB or BC's distance, since the train is stationary), or forg …