Answer to Question #91336 in Mechanics | Relativity for Deepak saini

Question

two railway station A and B and 50 kilometre apart and a electric train which can be accelerated 5 kilometre per hour per second and accelerated 3 km per hour per second the maximum speed is 90 km per hour there are 12 station all are more than a kilometre apart Find the least time which can be take to make a journey a to b.

Answer 1

The least time will be if our train moves with the acceleration $5\text{ km/h/s}$ . Calculate the distance necessary to accelerate the 🚆 from zero to its maximum speed:

$d=\frac{v_\text{max}^2}{2a}=\frac{(90\cdot1000/3600)^2}{2\cdot5\cdot1000/3600}=225\text{ m}.$

This will take time of

$t=v_\text{max}/a=90/5=42\text{ s}.$

The rest of the distance the train will do with its maximum speed:

$t_r=\frac{D-d}{v_\text{max}}=\frac{50000-225}{90\cdot1000/3600}=1991\text{ s}.$

The total time will be

$T=t+t_r=33.9\text{ min}.$