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Prove it

The speed of a train increases at a constant rate \(\huge\alpha\) from zero to V,and then remains constant for an interval, and finally decreases to zero at a constant rate \(\huge\beta\).if L be the total distance travelled , then the total time taken is given by:

\[\large \frac{L}{V}+ \frac{V}{2}(\alpha^{-1}+\beta^{-1})\]

Note by Rohit Udaiwal
2 years, 2 months ago

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A graphical approach might be best. Plot a graph of velocity versus time. The area under this graph represents the distance traveled. So starting at the origin, the first part of the graph is a line with slope \(\alpha\) terminating at the point \((V,t_{1}).\) The second part of the graph is a horizontal line going from \((V,t_{1})\) to \((V,t_{2}).\) The third and final part of the graph is a line with slope \(-\beta\) going from \((V,t_{2})\) to \((0, t_{3}).\)

Now the area under this graph is the total distance traveled \(L,\) and the quantity we want to find is \(t_{3}.\) Breaking the graph up into three sections, namely a rectangle flanked by two triangles, we see that

\(L = \dfrac{1}{2}V(t_{1} - 0) + V(t_{2} - t_{1}) + \dfrac{1}{2}V(t_{3} - t_{2}) \Longrightarrow \dfrac{L}{V} = \dfrac{1}{2}t_{1} + (t_{2} - t_{1}) + \dfrac{1}{2}(t_{3} - t_{2}),\) (i).

Next note that \(t_{1} = \dfrac{V}{\alpha}\) and that \(t_{3} - t_{2} = \dfrac{V}{\beta}.\) We then also have that

\(t_{2} - t_{1} = t_{3} - (t_{3} - t_{2}) - t_{1} = t_{3} - \dfrac{V}{\beta} - \dfrac{V}{\alpha}.\) Equation (i) then becomes

\(\dfrac{L}{V} = \dfrac{V}{2\alpha} + t_{3} - \dfrac{V}{\beta} - \dfrac{V}{\beta} + \dfrac{V}{2\beta} \Longrightarrow t_{3} = \dfrac{L}{V} + \dfrac{V}{2}\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\right)\) as required.

Brian Charlesworth - 2 years, 2 months ago

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you are the best sir !

Rohit Udaiwal - 2 years, 2 months ago

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Haha Thanks! Glad I could help. :)

Brian Charlesworth - 2 years, 2 months ago

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any help will be thankful \(\ddot\smile\)

Rohit Udaiwal - 2 years, 2 months ago

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