### Stopping Time for a Train

This problem belong to the group of problems where two equations must be used in order to obtain the main result. Part one of the problem corresponds to the main part of the problem. As you can see, after solving for the first part, the second part of the problem can be easily obtained.

A train is moving at 100 Km/H when the brakes are applied by the engineer, the train is stopped in 2 Km after the brakes are applied. Assuming that the deceleration of the train is constant,

What is the train stopping time in seconds?

 N a) 144 s b) 72 s c) 36 s d) 168 s e) None of the above.

Find the magnitude of the deceleration of the train.

 a) 0.39 m/s2 b) 0.77 m/s2 N c) 0.19 m/s2 d) 0.16 m/s2 e) None of the above.
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 ,

(U_A_M 5), is used. Still the first step will be to solve for the acceleration in formula (U_A_M 5),

Substituting the values of the known variables, (notice that with this method still is necessary to change the velocity to the standard unit of velocity)

Of course this result coincide with the result obtained using the first method.

 by Luis F. Sáez, Ph. D. Comments and Suggestions: LSaez@dallaswinwin.com