Please help! I am finding the new speed and distance traveled of train. Thanks!

  • Thread starter keevo
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  • #1
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1. Homework Statement

A train has a mass of 5.77E+6 kg and is moving at 99.4 km/hr. The engineer applies the brakes, which results in a net backward force of 1.11E+6 N on the train. The brakes are held on for 26.9 s.

What is the new speed of the train?

How far does it travel during this period?

mass of train: 5,770,000 kg

speed of train: 99.4 km/hr (i converted it to m/s and got 27.611 m/s)

Backwards net force of: 1,110,000 N

Brakes held for: 26.9 s

2. Homework Equations

f=ma?

V=Vo+at?

X=Xo+(Vo)(t)+.5at^2?

3. The Attempt at a Solution

I am fairly new to physics and don't really know what I'm doing, anyways...first i used f=ma
and got a=.192 then i used that in the second equation and got V=1,110,005.165 N and then I plugged that into the third equation and got 29859069.4666 M but it said i was wrong :/.

Can someone please guide me through this problem? it would be much appreciated

-Thanks in advance!

Ok, so i tried it again and i was confident that i was correct.

i used f=ma to get a=.192

i then used a in the second equation ( V=27.611+.192(26.9)) to get a Velocity of 32.7758

i then used the 3rd equation to get a distance of 812.20246 m but it said it was wrong again.

can you help me clarify or show me where i went wrong?

Thanks
 

Answers and Replies

  • #2
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^bump
 
  • #3
PhanthomJay
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If the train is braking, it is slowing down, so it's speed after the given elapsed time must be less than at the start. You are making an error in signage for the acceleration.
 
  • #4
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If the train is braking, it is slowing down, so it's speed after the given elapsed time must be less than at the start. You are making an error in signage for the acceleration.
so instead of .192 for Acceleration, would it be -.192?
 
  • #5
34,448
10,563
so instead of .192 for Acceleration, would it be -.192?
Right.

Why did you delete your post here?
 
  • #6
22,097
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1. Homework Statement

A train has a mass of 5.77E+6 kg and is moving at 99.4 km/hr. The engineer applies the brakes, which results in a net backward force of 1.11E+6 N on the train. The brakes are held on for 26.9 s.

What is the new speed of the train?

How far does it travel during this period?

mass of train: 5,770,000 kg

speed of train: 99.4 km/hr (i converted it to m/s and got 27.611 m/s)

Backwards net force of: 1,110,000 N

Brakes held for: 26.9 s

2. Homework Equations

f=ma?

V=Vo+at?

X=Xo+(Vo)(t)+.5at^2?

3. The Attempt at a Solution

I am fairly new to physics and don't really know what I'm doing, anyways...first i used f=ma
and got a=.192 then i used that in the second equation and got V=1,110,005.165 N and then I plugged that into the third equation and got 29859069.4666 M but it said i was wrong :/.

Can someone please guide me through this problem? it would be much appreciated

-Thanks in advance!

Ok, so i tried it again and i was confident that i was correct.

i used f=ma to get a=.192

i then used a in the second equation ( V=27.611+.192(26.9)) to get a Velocity of 32.7758

i then used the 3rd equation to get a distance of 812.20246 m but it said it was wrong again.

can you help me clarify or show me where i went wrong?

Thanks
so instead of .192 for Acceleration, would it be -.192?
Can people please quote everything so we can restore threads when necessary? Thanks!
 

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