Question

A train starts from rest and its speed increases at a constant rate $\alpha$ to $v$ and then remains constant for an interval and finally decreases to zero at a constant rate $\beta$. The total distance travelled by 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. $\text{t is minimum when v}=\sqrt{\frac{2t\alpha \beta }{(\alpha -\beta )}}$
• D. $\text{t is minimum when v}=\sqrt{\frac{2l\alpha \beta }{(\alpha +\beta )}}$

Answer 1

Answer: (B), (D)

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

At $t=0$, $u=0$

From the figure
$t=t_1+t_2+t_3\left(1\right)$
By using first equation of motion, we get
$t_1=\frac{v}{a}\text{and}{t}_3=\frac{v}{\beta }\left(2\right)$
Let $l_2$ be the distance covered with constant velocity
$\Rightarrow t_2=\frac{l_2}{v}\left(3\right)$
By using third equation of motion:
also , $l_1=\frac{v^2}{2\alpha }\text{and}{l}_3=\frac{v^2}{2\beta }$
According to question,
$l=l_1+l_2+l_3$
$\therefore l_2=l-\frac{v^2}{2\alpha }-\frac{v^2}{2\beta }\left(4\right)$
Using equation (1), (2) and (3) we get
$t=\frac{v}{\alpha }+\frac{v}{\beta }+\frac{l_2}{v}\left(5\right)$
Using (4) and (5), we get
$t=\frac{v}{\alpha }+\frac{v}{\beta }+\frac{l}{v}-\frac{v}{2\alpha }-\frac{v}{2\beta }$
$\therefore t=\frac{l}{v}+\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$
For $t$ to be minimum
$\frac{dt}{dv}=0$
$-\frac{l}{v}+\frac{1}{2}\left(\frac{\beta +\alpha }{\alpha \beta }\right)=0$
$\Rightarrow v=\sqrt{\frac{2l\alpha \beta }{(\alpha +\beta )}}$