Steven W. answered • 10/09/16
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Hi Nona!
If we start our clock at the moment a spot on Train A passes Train B, we can write the displacements of the two trains as functions of time:
dA = vAt (traveling at constant velocity)
dA = 20t
dB = voB(t)+(1/2)(aB)t2 = 5t + (1/2)(0.5)t2 = 5t+0.25t2
To solve for the time when the trains are again beside each other, we can set the two displacements equal, then solve for the value of t that makes that true.
dA = dB
20t = 5t + 0.25t2
0.25t2-15t = 0
t(0.25t-15) = 0
So, one of the solutions for when the trains are side-by-side is t = 0, which we already know, because they were beside each other when we started our clock. The other time is when
0.25t-15 = 0 --> 0.25t = 15 --> t = 60 s
You can then plug this time back into either the dA or dB equations to solve for the displacement at the time they meet again.
I hope this helps out!
Nona Z.
Mr.Steven thanks a lot for your help ,I've already finished my Mechanics lesson ,I noticed a very big difference in the way that I'm thinking, even the teacher was really happy , I really deticate this for yourself as you have helped me a lot .
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10/10/16
Steven W.
tutor
Very good! Remember that the key is not just in getting the right answers, but understanding the process. If you can do that, you can deal with a range of problems in kinematics (and other topics) that you may not have encountered yet.
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10/10/16
Nona Z.
10/10/16