Problem 11 - Train acceleration and deceleration
A train accelerates at 0.9m/s2 for 500m from the start and then decelerates at 0.9m/s2 on the next 500m. What is the travel time? What is its maximum speed?
Given:
D1 = D2 = D = 500m – half of the total distance traveled,
a1 = 0.9 m/s2 – acceleration during the first half of the travel time,
a2 = -0.9m/s2 – acceleration during the second half of the travel time,
We are looking for:
t = ? – total time of travel
v1 = ? – maximum speed.
To find t and v1 we must calculate:
t1 = ? – time needed to travel the first 500m,
t2 = ? – time needed to travel the second 500m.
Analysis.
Intuition tells us that t1 = t2, but we will prove this rigorously. To simplify equations used to solve this problem we will only use one symbol a for acceleration and a plus or minus sign will be used to distinguish acceleration from deceleration.
We write three equations based on laws of linear motion
D = (1/2)at12 (1)
D = v1t2 – (1/2)at22 (2)
v1 = at1 (3)
From Eq.1
(4)
Substituting this into Eq. 3 we get the speed after traveling the first 500m, that is the maximum speed, as from this moment train starts deceleration.
(5) - this
is maximum speed
Now we substitute Eq.2 and Eq.5 into Eq.2
(6)
This can be rearranged to the form
(7)
This is a well known from algebra quadratic equation of the type
Ax2 + Bx + C = 0,
with
, 
, and C = D
Solving Eq.7 gives two possible values for t2, (quadratic equation !)
(8)
(9)
We reject the negative value, as the time duration cannot be negative.
Time given by Eq.8 is the same as time given by Eq.4, so we proved what seems to us intuitively obvious. The final answer for time of travel t is
t = t1 + t2
or
