Question

A subway train stops at two stations that are 2 km. apart. The maximum Acceleration and deceleration of the train are 2 m/s^2 and 1.6 m/s^2 Respectively and the maximum allowable speed is 20 m/s.

1. What is the time to reach a max. velocity.

2. Which of the following gives the time to travel constant speed of 20 m/s.

3. Which of the following gives the shortest possible time of travel between two stations.

Answer 1

1. Initial speed = 0

Acceleration a = 2 m/s²

Time to reach v = 20 m/s

Using

$v=u + at \\ t = \frac{20-0}{2} = 10 \; sec$

2. Distance covered in this time = S₁

$S_1 = ut + \frac{1}{2}dt^2 \\ = \frac{1}{2} \times 2 \times 10^2 \\ = 100 \;m$

Time needed to decelerate from 20 m/s to 0 m/s

a = -1.6 m/s²

$v=u+at \\ 0 = 20 -1.6t \\ t = 12.5 \;sec$

3. Distance covered in this time

$S_3 = ut +\frac{1}{2}at^2 \\ = 20 \times 12.5 -\frac{1}{2} \times 1.6 \times 12.5^2 \\ = 125 \;m \\ S_2 = 1000 -100 -125 = 775 \;m \\ t_2 = \frac{S_2}{20} \\ t_2 = 38.75 \;sec$