How long does it take for the train to stop moving?

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In summary, the train is traveling at 20 m/s on a straight track and the engineer applies the brakes resulting in an acceleration of -1.0 m/s^2. The train stops moving after 20 seconds, and therefore travels a total of 210 meters during the 40 second time interval. The equation used to calculate this distance is 20+19+18+...+1+0=210 meters. Common sense can be used to understand the problem, but it is important to also use the kinematic relationships to fully comprehend the concept.
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blader324
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Homework Statement


a train is traveling down a straight track at 20m/s when the engineer applies the brakes resulting in an acceleration of -1.0m/s^2 as long as the train is in motion. how far does the train move durin g a 40 second time interval starting at the instant the brakes are applied?


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The Attempt at a Solution


I'm new to physics and have no idea what to do. i keep getting zero when i plug it into the formula. also...i know that the train eventually stops before 40 seconds and doesn't travel anywhere...but how do i incorporate that into the equation?
 
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  • #2
How long does it take for the train to stop moving?
 
  • #3
i think it takes 20 seconds...but I'm not so sure.
 
  • #4
That's correct. If you're not sure, review the definition of acceleration.
 
  • #5
okay...so if it takes 20 seconds for the train to stop moving...doesn't that mean it travels for a total of 210 meters?
 
  • #6
How did you deduce that? (Always show your work.)
 
  • #7
i'm using common sense for the problem but i don't know what equation to use to get the answer of 200 meters. i just did 20+19+18+...+1+0=210 meters.
 
  • #8
blader324 said:
i'm using common sense for the problem but i don't know what equation to use to get the answer of 200 meters. i just did 20+19+18+...+1+0=210 meters.
Commonsense is OK if you can justify it. What do those numbers mean? 20 = ?; 19 = ?...

If you are thinking that the train travels 20 m in the first second, etc., that's not quite correct. Hint: What's the average speed during the first second?

I suspect you should be learning some kinematic relationships.
 
  • #9
thanks so much...i figured it out...and finally understand what I'm doing with all my other homework problems!
 

1. How is the stopping distance of a train calculated?

The stopping distance of a train is calculated using the formula: stopping distance = initial velocity² / (2 x deceleration). The initial velocity is the speed of the train before it starts to slow down, and the deceleration is the rate at which the train slows down.

2. How long does it take for a train to come to a complete stop?

The time it takes for a train to come to a complete stop depends on several factors, such as the train's initial speed, weight, and braking system. On average, a train traveling at 60 mph can take up to 1 mile to come to a complete stop.

3. What factors affect the stopping distance of a train?

The stopping distance of a train can be affected by various factors, including the train's speed, weight, length, braking system, and weather conditions. Other factors like the condition of the track and the grade of the track can also impact the stopping distance.

4. How does the length of the train impact its stopping distance?

The length of a train can have a significant impact on its stopping distance. A longer train will have more momentum and will take longer to slow down compared to a shorter train. This is why it's important for trains to maintain a safe distance between each other to allow for enough stopping distance.

5. Can a train stop faster if the brakes are applied harder?

Yes, a train can stop faster if the brakes are applied harder. The braking system of a train is designed to exert a certain amount of force to slow down the train. The harder the brakes are applied, the more force is exerted, resulting in a shorter stopping distance. However, applying the brakes too hard can cause the train to slide or derail, so it's important for the train operator to use the appropriate amount of braking force.

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