# Determine stopping distance of a train traveling at 20m/s

• Alexanddros81
In summary, a train traveling at 20m/s is brought to an emergency stop with an acceleration of a=-(7/4)+(t/16) m/s^2. Using Euler's method with Δt=2s, the stopping distance of the train is between 150.5-152 m, which is above the analytically found value of 138.7m. Linear interpolation can be used to find the stopping distance between x7 and x8, but it will not significantly increase the accuracy due to the error already present in the Euler integration method.
Alexanddros81
This is problem 12.81 from Pytels Dynamics 2nd edition

1. Homework Statement

A train traveling at 20m/s is brought to an emergency stop. During braking,
the acceleration is a=-(7/4)+(t/16) m/s^2, where t is the time in seconds measured
from when the brakes were applied. (a) Integrate the acceleration from t=0 to
t=16s using Euler's method with Δt=2s. (b) Use the results of the integration to
determine the stopping distance of the train and compare you answer with 138.7m,
the value found analytically.

## The Attempt at a Solution

[/B]
According to the results of the Euler method above the stopping distance is between 150.5-152 m
That is way above 138.7m. Am I correct?

Looks good to me.

Is the stopping distance between x7 and x8 where the velocity is 0?

Do I use linear interpolation to find the stopping distance?
for example:

Do I use something like this?

If yes how do I apply linear interpolation in above graph to find x?

You could do that, but as you see there is little gain in accuracy (*)
You would use something like $${x-150\over v_7} = {x-152\over v_8} \Rightarrow x = \left ( {150.5\over v_7}-{152\over v_8} \right )/ \left ({1\over v_7}-{1\over v _8}\right )$$ or 151.14.

(*) the error the Euler integration makes in the first step is already about 3.5 m
( ## x = v_0 t -{1\over 2} a_0 t^2 \ ## gives 36.5 m instead of 40 m, the exact formula ##v_0 t - {1\over 2} {7\over 4} t^2 + {1\over 2}{1\over 3} {\displaystyle t^3\over 16} ## gives 36.6 m )

## 1. How do you calculate the stopping distance of a train traveling at 20m/s?

The stopping distance of a train can be calculated using the formula: stopping distance = (initial velocity)^2 / (2 * deceleration), where deceleration is the rate at which the train slows down. In this case, the initial velocity is 20m/s, so the stopping distance would be (20)^2 / (2 * deceleration).

## 2. What factors affect the stopping distance of a train?

The two main factors that affect the stopping distance of a train are the initial velocity and the deceleration rate. Other factors that can also have an impact include the weight and length of the train, the condition of the brakes, and the surface of the track.

## 3. How does the deceleration rate of a train affect its stopping distance?

The deceleration rate of a train is directly proportional to its stopping distance. This means that the higher the deceleration rate, the shorter the stopping distance will be. On the other hand, a lower deceleration rate will result in a longer stopping distance.

## 4. Can the stopping distance of a train be affected by external factors?

Yes, external factors such as weather conditions, track conditions, and the weight and length of the train can all impact the stopping distance. For example, a wet or icy track will increase the stopping distance, while a heavier train will require a longer distance to come to a complete stop.

## 5. How can the stopping distance of a train be reduced?

The stopping distance of a train can be reduced by increasing the deceleration rate, improving the condition of the brakes, and keeping the weight and length of the train to a minimum. Proper maintenance of the track and following speed limits also play a crucial role in reducing the stopping distance of a train.

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