- #26

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One of the problems is that you haven't shown us any of your work yet. We've no idea how you are trying to solve this problem, what your algebra skills are like and how you use the kinematic equations.May I ask something please?

How can I insert time(n) into those kinematic equations?

I would like to solve this problem as fast as it could be but comparing time(1) and time(2) and etc. Isn't fast enough.

I hope you won't get angry because of my question.

Thank You

For example, suppose you were asked to find the speed of the train after the nth wagon has passed, then I would do something like:

Let ##L## be the length of each wagon.

We have the kinematic equation: ##v^2 - u^2 = 2as##, which in this case (##u = 0##) reduces to ##v^2 = 2as##, where ##a## is the constant acceleration of the train.

Let ##v_n## be the speed of the train after ##n## wagons have passed. Then we have:

##v_n^2 = 2a(nL)##

And, suppose you were given that the speed of the train after the first wagon was ##v_1 = 2m/s##, and the length of each wagon is ##16m##. Note: this is a different problem from your one. Then:

##v_1^2 = 2aL##

##a = \frac{v_1^2}{2L} = \frac{4m^2/s^2}{32m} = 0.125 ms^{-2}##

And, finally,

##v_n^2 = 2(0.125 ms^{-2})n(16m) = (4n)m^2s^{-2}##

##v_n = (2\sqrt{n}) ms^{-1}##

That's not necessarily the quickest way to solve the problem, but I wanted to illustrate the way you should be approaching these problems. How to introduce variables, and use and solve the kinematic equations.