PYQ NEET- Motion In A Straight Line Kinematics L-4

Question: Preeti reached the metro station and found that the escalator was not working. She walked up the sationary escalator in time $t_1$. On another days, if she remains stationary on the moving escalator, then the escalator takes her up in time $t_2$. The time taken by her to walk up on the moving escalator will be

A) $\frac{t_1 t_2}{t_2-t_1}$

B) $\frac{t_1 t_2}{t_2+t_1}$

C) $t_1-t_2$

D) $\frac{t_1+t_2}{2}$

Answer: $\frac{t_1 t_2}{t_2+t_1}$

Sol:

Velocity of preeti with respect to elevator $\mathrm{v}_1=\frac{d}{t_1}$ Velocity of elevator with respect to ground $\mathrm{v}_2=\frac{d}{t_2}$ $\therefore$ Net velocity of preeti on moving escalator with respect to the ground $$ \begin{aligned} & \mathrm{v}=\mathrm{v}_1+\mathrm{v}_2 \ & \frac{d}{t}=\frac{d}{t_1}+\frac{d}{t_2} \ & \frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2} \ & \therefore \mathrm{t}=\frac{t_1 t_2}{t_1+t_2} \end{aligned} $$

Here $t$ is the time taken by preeti to walk up on the moving escalator.



sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language