Question

A train accelerates at 1.41 m/s² from rest to a steady speed in 68.48seconds. It maintains that speed for 29.55 miles and then starts slowing down at 1.39 m/s² until coming to a stop at its destination. What is the total distance traveled in meters?

Answer 1

Here first train will accelerate with a = 1.41 m/s^2

Initial velocity V₁ = 0 m/s so it start with rest and accelerate for time 68.48 sec

t = 68.48 sec and distance covered in this time d₁

We can use equation d = Vo*t + 1/2a*t^2

First we will calculate distance d₁using the same equation

d₁ = V₁*t + 1/2*a*t^2

d₁ = (0)*68.48 + 1/2 (1.41)(68.48)^2

d₁ = 3306.10 meter

and its final velocity after this time is V₂

V₂ = V₁ + a*t

V₂ = 0 + (1.41)*68.48

V₂ = 96.56 m/s

Then it cover 29.55 miles with constant speed we will change it in meter 1 mile = 1609.344 m

d₂ = 29.55 miles = 29.55 (1609.344 meter)

d₂ = 47556.11 m

now it will start decelerating (slowing down) with an acceleration of a' = -1.39 m/s^2

we can use V₃^2 = V₂^2 + 2*a'*d₃ velocity from which it start slwoing down and final velocity V₃=0

It was moving with 96.56 m/s and then slowing down

0 = (96.56)^2 + 2*(-1.39)(d₃)

0 = 9323.22 - 2.78 d₃

2.78 d₃=9323.22

d₃ = 3353.67 m

So total distance D= d1+d2+d3

D = 3306.10 + 47556.11 +3353.67

D = 54215.88 m