A train starts from rest at station A and accelerates at 03

A train starts from rest at station A and accelerates at 0.3 m/s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 0.6 m/s2 until it is brought to rest at station B. Determine the distance between the stations. s =

Solution

Case 1: At Station A, as Train is at rest, Initial Velocity of train = 0

Given acceleration = 0.3 m/s² for time of 60 secs

Using the formula, s = ut+1/2at²

=> s = (0 * 60) + (1/2 * 0.3 * 60²)

on solving, => s = 540m

And also using the formula, v = u + at

=> v = 0 + (0.3 * 60)

Hence final velocity => v = 18 m/s

Case 2: In this case, Train travels for 15 min (900 secs) with constant velocity

And constant velocity will be final velocity in previous case i.e. = 18 m/s and so acceleration = 0

Using s =ut + 1/2at² and substituting u = 18, t = 900, a = 0, we get s = 16200

Case 3: In thi case, as train decelerates a = -0.6 m/s2 and initial velocity = 18 m/s and final velocity = 0

Using v² - u² = 2as and substituting v = 0, u = 18, a = -0.6. we get s = 270m

Now distance between stations A and B = 540 + 16200 +270 = 17010m