The speed of train increases at a constant rate alfa from zero to V and then remains constant for an interval and finally decreses to zero at a constant rate Beta. If L be the total distance travelled then total time taken is?

Answers (1)

first velocity increases from 0 to V,

So 

use \ first \ equation\\ \mathbf{v}=\mathbf{u}+\mathbf{a}* \mathbf{t}: \mathbf{u}=\mathbf{0}, \mathbf{v}=\mathbf{V}, \mathbf{a}=\mathbf{\alpha }$ this gives $t=\frac{V}{\alpha}$\\ use second equation\\ $\mathbf{s}=\mathbf{u}* \mathbf{t}+\frac{1}{2} * a * t^{2}: \mathbf{u}=\mathbf{0}, \mathbf{a}=\mathbf{a}, \mathbf{t}=\frac{V}{\alpha}$ this gives $\mathbf{s_1} =\frac{V^{2}}{2 * \alpha}$

then velocity decreases from V to 0,

So

use \ first \ equation \\ \ \ \mathbf{v}=\mathbf{u}+\mathbf{a}*\mathbf{t}: \mathbf{u}=\mathbf{0}, \mathbf{v}=\mathbf{V}, \mathbf{a}=-\mathbf{\beta}$ \\ this gives $t=\frac{V}{\beta}$\\ use second equation $s=u*t+\frac{1}{2} * a * t^{2}: u=V, a=-\beta, t=\frac{V}{\beta}$ \\ this gives $\mathbf{s_2}=\frac{V^{2}}{2 * \beta}$\\ So total distance traveled $\mathbf{d}=\frac{V^{2}}{2} *\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$

Thus distance traveled during constant velocity phase 

 =L-(\frac{V^{2}}{2} *(\frac{1}{\alpha}+\frac{1}{\beta}))

So time taken to travel this distance 

\mathbf{T}=\frac{L-\frac{V^{2}}{2} *\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)}{V} \Rightarrow \frac{L}{V}-\frac{V}{2} *\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)

So total time taken to cover total distance L

 

=$ $\left(\frac{V}{\alpha}\right)+\left(\frac{L}{V}-\frac{V}{2} *\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\right)+\left(\frac{V}{\beta}\right)$ $\Rightarrow \frac{L}{V}+\frac{V}{2} *\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$

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