Asubway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. it runs at constant speed for 68.0 s and slows down at a rate of 3.70 m/s2 until it stops at the next station. what is the total distance covered in kilometers?

1868.5 m

Explanation:

For AB :

u = 0 m/s

a = + 1.68 m/s^2

t = 14.2 s

Let the distance is s1 and the velocity at B is v.

Use first equation of motion

v = u + at

v = 0 + 1.68 x 14.2 = 23.856 m/s

Use third equation of motion

s1 = 169.38 m

For BC:

Let the distance is s2.

s2 = v x t

s2 = 23.856 x 68 = 1622.21 m

For CD:

u = 23.856 m/s

a = - 3.7 m/s^2

v = 0

Let the distance is s3.

Use third equation of motion

s3 = 76.91 m

The total distance traveled is

s = s1 + s2 + s3

s = 169.38 + 1622.21 + 76.91 = 1868.5 m

Thus, the total distance traveled is 1868.5 m.  

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more