Calculating Travel Time for Commuter Train

  • Thread starter Thread starter miscellaneous
  • Start date Start date
  • Tags Tags
    Time Train Travel
Click For Summary

Homework Help Overview

The discussion revolves around calculating the total travel time for a commuter train between two stations, focusing on the effects of acceleration and deceleration. The problem involves kinematic equations and the train's motion over a distance of 0.90 km.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to break the motion into segments with different accelerations. There is an emphasis on confirming the calculations and understanding the relationship between initial, final velocities, and acceleration.

Discussion Status

Some participants are seeking validation of their methods and calculations, while others express uncertainty about the correctness of their approaches. There is a general exploration of the problem without a clear consensus on the final answer.

Contextual Notes

Participants note that the initial and final speeds are both zero over the entire distance, which may influence their calculations. There is a focus on ensuring that the kinematic equations are applied correctly for each segment of the train's motion.

miscellaneous
Messages
10
Reaction score
0
A commuter train can minimize the time t between two stations by accelerating (a1 = .20 m/s^2) for a time t1, then undergoing a negative acceleration (a2 = -.5 m/s^2) by using his brakes for a time t2. Since the stations are .90 km apart, the train never reaches its maximum velocity. Find the total travel time, t.

My method:

deltaS= (Vf^2-Vo^2)/2a
900= (Vf^2/.4) + (Vf^2)
900= 2.5Vf^2 + Vf^2
Vf= 16 m/s

and then I used: Vf = Vo + at
16= .2t
t=80 s

0= 16 - .5t
t=32 s

total t= 112 seconds

Can anyone please confirm this?
 
Physics news on Phys.org
miscellaneous said:
My method:

deltaS= (Vf^2-Vo^2)/2a
900= (Vf^2/.4) + (Vf^2)
900= 2.5Vf^2 + Vf^2
Vf= 16 m/s
900 m is the entire distance. Note that the initial and final speeds are both zero over the 900 m.

Break the motion into two pieces, each with a different acceleration. Use that same kinematic formula, but for each piece separately. (The first piece goes from v = 0 to v = V1; the second goes from v = V1 to v = 0.)
 
I think that's what I did. Is my answer incorrect? I'm pretty sure my method is what you explained. If I'm wrong, please let me know. The Vf is the same as V1
 
miscellaneous said:
I think that's what I did. Is my answer incorrect? I'm pretty sure my method is what you explained. If I'm wrong, please let me know. The Vf is the same as V1
Looks good. (I missed what you did, at first look.)
 
Ohh, thanks for the confirmation. I'm really happy I know what I'm doing :approve:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
Replies
2
Views
2K
Replies
5
Views
7K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
7K