Question

On a linear escalator running between two points $A$ and $B$, a boy takes time $t_1$, to move from $A$ to $B$, if the boy runs with a constant speed on the escalator. If the boy runs from $B$ to $A$, he takes time $t_2$ to reach $B$ from $A$. The time taken by the boy to move from $A$ to $B$ if he stands still on the escalator will be (The escalator moves from $A$ to $B$)

• A. $\frac{t_1{t}_2}{t_2-t_1}$
• B. $\frac{{2t}_1{t}_2}{t_2-t_1}$
• C. $\frac{t_1^2+t_2^2}{t_1{t}_2}$
• D. $\frac{t_1^2-t_2^2}{t_1{t}_2}$

Answer 1

The correct option is B $\frac{{2t}_1{t}_2}{t_2-t_1}$

Let the distance between A and B=$d$, Speed of boy = $u$ and speed of escalator = $v$.

Let time $t$ taken by boy if he stands still on escalator.

Relative speed of boy when he moves from A to B= $u+v$,

Relative speed of boy when he moves from B to A= $u-v$,

$\therefore t_1=\frac{d}{u+v}$ $\Rightarrow u+v=\frac{d}{t_1}$ ...1

$\therefore t_2=\frac{d}{u-v}$ $\Rightarrow u-v=\frac{d}{t_2}$ ...2

Subtracting equation 2 from 1, $\Rightarrow 2v=d(\frac{1}{t_1}-\frac{1}{t_2})$ $\Rightarrow \frac{d}{v}=\frac{2t_1{t}_2}{t_2-t_1}$ ....3

$\therefore t=\frac{d}{v}=\frac{2t_1{t}_2}{t_2-t_1}$