Commuter train

  • #1
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?
 

Answers and Replies

  • #2
Doc Al
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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.)
 
  • #3
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
 
  • #4
Doc Al
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44,953
1,220
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.)
 
  • #5
Ohh, thanks for the confirmation. I'm really happy I know what I'm doing :approve:
 

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