Question

A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s 2 for 14.0 s. It runs at constant speed for 70.0 s and slows down at a rate of 3.50 m/s 2 until it stops at the next station. Find the total distance covered.

Answer 1

1796.48 m

Step-by-step explanation:

Given that :

First part of journey :

Initial Velocity, u = 0

Acceleration, a, = 1.60 m/s²

Time, t = 14 s

Distance traveled, S = 0.5at²

S = 0.5 * 1.60 * 14²

S1 = 156.8m

2nd part :

Speed is constant

Time = 70 seconds

At constant speed ;

Distance = speed * time

Speed, V = u + at

V = 0 + 1.6*14

V = 22.4 m/s

Distance, S2 = 22.4 * 70 = 1568 m

3rd part :

Deceleration = - 3.50m/s²

Final velocity, v = 0

Time taken to attain rest

V = u + at

0 = 22.4 - 3.5(t)

3.5t = 22.4

t = 22.4/3.5

t = 6.4 seconds

S3 = ut - 0.5at² (deceleration)

S3 = (22.4*6.4) - 0.5(3.5)*6.4^2

S3 = 71.68m

S1 + S2 + S3

156.8m + 1568m + 71.68m

= 1796.48 m