Question

The speed of a train increases at a constant rate $\alpha$ from zero to $v$ and then remains constant for an interval and finally decreases to zero at a constant rate $\beta$. The total distance travelled by the train is $l$. The time taken to complete the journey is $t$. Then

• A. $t=\frac{l(\alpha +\beta )}{\alpha \beta }$
• B. $t=\frac{l}{v}+\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$
• C. $t$ is minimum when $v=\sqrt{\frac{2l\alpha \beta }{(\alpha -\beta )}}$
• D. $t$ is minimum when $v=\sqrt{\frac{2l\alpha \beta }{(\alpha +\beta )}}$

Answer 1

Answer: (B), (D)

The correct options are
B $t=\frac{l}{v}+\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$
D $t$ is minimum when $v=\sqrt{\frac{2l\alpha \beta }{(\alpha +\beta )}}$

$v=\alpha t_1\Rightarrow t_1=\frac{v}{\alpha }$
$v=\beta t_1\Rightarrow t_2=\frac{v}{\beta }$
$\therefore t_0=t-t_1-t_2=(t-\frac{v}{\alpha }-\frac{v}{\beta })$
$Now,l=\frac{1}{2}\alpha t_1^2+v{t}_0+\frac{1}{2}\beta t_2^1$

$\Rightarrow \frac{1}{2}\alpha {\left(\frac{v}{\alpha }\right)}^2+v(t-\frac{v}{\alpha }-\frac{v}{\beta })+\frac{1}{2}\beta {\left(\frac{v}{\beta }\right)}^2\Rightarrow vt-\frac{v^2}{2\alpha }-\frac{v^2}{2\beta }\Rightarrow t=\frac{l}{v}+\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$
For $t$ to be minimum its first derivative with respect to velocity should be zero or,

$\Rightarrow 0=-\frac{l}{v^2}+\frac{\alpha +\beta }{2\alpha \beta }\Rightarrow v=\sqrt{\frac{2l\alpha \beta }{\alpha +\beta }}$