Question

A subway 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?

Answer 1

1868.5 m

Step-by-step 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

`v^(2)=u^(2)+2as_(1)`

`23.856^(2)=0^(2)+2*1.68* s_(1)`

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

`v^(2)=u^(2)+2as_(3)`

`0^(2)=23.856^(2)-2* 3.7 * s_(3)`

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.

Answer 2

total distance = 1868.478 m

Step-by-step explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at² ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m