At a metro station, a girl walks up a stationary escalator in time t1 . If she remains stationary on the escalator, then the escalator take her up in time t2. The time taken by her to walk up on the moving escalator will be

The correct answer is $\frac{t_1{t}_2}{t_2+t_1}$.

To find the time taken by the girl to walk up on the moving escalator, let us analyze the speeds in each scenario. Let the total length of the escalator be represented as $L$.

Step 1: Find the walking speed of the girl
When the escalator is stationary, the girl walks up the distance $L$ in time $t_1$. Her walking speed $v_1$ is:
$v_1=\frac{L}{t_1}$

Step 2: Find the speed of the escalator
When the girl remains stationary, the moving escalator carries her up the distance $L$ in time $t_2$. The speed of the escalator $v_2$ is:
$v_2=\frac{L}{t_2}$

Step 3: Find the combined speed on the moving escalator
When the girl walks up the moving escalator, both she and the escalator move in the same direction. Her effective speed $v$ with respect to the ground is the sum of her walking speed and the escalator's speed:
$v=v_1+v_2$
Substituting the expressions for $v_1$ and $v_2$:
$v=\frac{L}{t_1}+\frac{L}{t_2}=L(\frac{1}{t_1}+\frac{1}{t_2})=L\left(\frac{t_2+t_1}{t_1{t}_2}\right)$

Step 4: Calculate the total time taken
The time $t$ taken to cover the distance $L$ at the combined speed $v$ is:
$t=\frac{L}{v}$
Substituting the expression for $v$:
$t=\frac{L}{L\left(\frac{t_2+t_1}{t_1{t}_2}\right)}=\frac{t_1{t}_2}{t_2+t_1}$