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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, 5 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.

- 2 years, 5 months ago

you are the best sir !

- 2 years, 5 months ago

Haha Thanks! Glad I could help. :)

- 2 years, 5 months ago

any help will be thankful $$\ddot\smile$$

- 2 years, 5 months ago