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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