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:

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

#Mechanics

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 5 years, 10 months ago

you are the best sir !

Rohit Udaiwal - 5 years, 10 months ago

Haha Thanks! Glad I could help. :)

Brian Charlesworth - 5 years, 10 months ago

any help will be thankful $\ddot\smile$

Rohit Udaiwal - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$