Question

Two boys enter a running escalator on the ground floor of a shopping mall and they do some fun on it. The first boy repeatedly follows $p_1=1$step up and then $q_1=2$ steps down whereas the second boy repeatedly follows $p_2=2$ steps up and $q_2=1$ step down. Both of them move relative to the escalator with speed $v_r=50cm{s}^{-1}$. If the first boy takes $t_1=250s$ and the second boy takes $t_2=50s$ to reach the first floor, how fast is the escalator running?

Answer 1

Hint: Students can apply the idea that the boy moving more steps down will take more time to reach the destination. And then using the very basic formula for speed, we can determine the velocity of the escalator.

Complete step by step solution:
Let us consider that the speed of the escalator is $v$.
We know the very basic formula,
$\begin{array}{l}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}={\displaystyle \frac{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}}{\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}}}\\ \Rightarrow \mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}\times \mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}=\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\end{array}$
In the given question, both the boys move the same distance, but there is a difference in the time taken by each one of them to reach the destination.
The first boy takes only one step up and two steps down, so naturally he is going to take more time than the second boy who is taking two steps up and only one step down.
Let the displacement while taking one step is $x$
So, the time taken to cover one step is ${\displaystyle \frac{x}{50}}$
In both the cases, the time taken for covering the steps either backward or forward will be ${\displaystyle \frac{3x}{50}}$.
However, for the first boy the total distance will be $-x$ ( one step up, two steps down ) and for the second boy, the total distance will be $x$.
So, for the first boy, his average speed should be ${\displaystyle \frac{-x}{{\displaystyle \frac{3x}{50}}}}=-{\displaystyle \frac{50}{3}}$.
Similarly for the second boy, his average speed should be ${\displaystyle \frac{x}{{\displaystyle \frac{3x}{50}}}}={\displaystyle \frac{50}{3}}$
So,
$\begin{array}{l}(v_r-{\displaystyle \frac{50}{3}})\times 250=(v_r+{\displaystyle \frac{50}{3}})\times 50\\ \Rightarrow 5\times (v_r-{\displaystyle \frac{50}{3}})=(v_r+{\displaystyle \frac{50}{3}})\\ \Rightarrow 4v_r=100\\ \Rightarrow v_r=25m/s\end{array}$
The speed of the escalator is $25m/s$.

Note:
Students must take care of the signs while calculations or else it will give us a different result. In case the first boy would have taken two steps up and one step down like the second boy, then his velocity would also have been the same as the second boy.