Solution:
Part 1:
The stopping time can be calculated from
, (U_A_M 1).
However, in order to utilize this formula, the average velocity of the
train need to be calculated. Since the deceleration of the train is
constant, the formula
, (U_A_M 2),
can be used for obtaining the average velocity. In order to use this
formula, the original and final velocity has to be identified from the
redaction of the problem. Since the train is moving at 100 Km/H, this
velocity correspond to the original velocity. In the same form, the
relevant time is the time when the train is completely stopped which
correspond to zero final velocity. Thus,
Now, at formula (U_A_M 1), the time is the
unknown quantity,
However, the question requires the answer in seconds, with
1H = 3600 s. Therefore,

Solution:
Part 2; Method 1, using the result of part 1:
Now that the time required to stop the train is known, formula
, (U_A_M 3),
can be solved for the acceleration (deceleration). Just as in part 1,
the original velocity is 100 Km/H and the final velocity is zero. The
time is 144 s. To obtain the acceleration in the standard units of
acceleration, the original velocity, 100 Km/H, must be changed to the
standard units of velocity,
Solving for the acceleration in formula (U_A_M 3),
Substituting the values of the known variables,
The previous result is the deceleration of the train
when the brakes are applied. The negative sign of the acceleration is a
remainder of the deceleration nature of this acceleration.
Method 2: This
calculation can also be obtained without using the pre calculated time
when the formula
, 