Question

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time $t_1$. On other 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 1

The Correct Option is B

Velocity of girl w.r.t. elevator $=\frac{d}{t_{1}}=v_{g e}$
Velocity of elevator w.r.t. ground $v_{e G}=\frac{d}{t_{2}}$ then
velocity of girl w.r.t. ground
$\vec{v}_{g G}=\vec{v}_{g e}+\vec{v}_{e G}$
i.e, $v_{g G}=v_{g e}+v_{e G}$
$\frac{d}{t}=\frac{d}{t_{1}}+\frac{d}{t_{2}}$
$\frac{1}{t}=\frac{1}{t_{1}}+\frac{1}{t_{2}}$
$t=\frac{t_{1} t_{2}}{\left(t_{1}+t_{2}\right)}$

Answer 2

Length of the path that is taken by the elevator be l

Let the speed of Preeti’s walking be vP and the speed of the elevator be v_E

It is given the time taken by Preeti to walk up the length of the stationary escalator t₁.

Also, the time taken by the elevator to go up at length with Preeti stationary on it is t_2.

Now,

Speed = \(\frac{Distance}{Time}\) ………(1)

Therefore, using (1), we get,

v_P=\(\frac{l}{t_1}\)………(2)

v_E=\(\frac{l}{t_2}\)……..(3)

As explained, when she walks up along the moving escalator, her total speed 

v will be the total speed of the elevator and her individual speed.

Hence,

v=v_P+v_E

Using (2) and (3), we get,

v=\(\frac{l}{t_1}+\frac{l}{t_2}\)= \(l(\frac{1}{t_1}+\frac{1}{t_2})\)=\(l(\frac{t_2+t_1}{t_1 t_2})\) = \(\frac{l}{(\frac{t_1t_2}{t_2 + t_2})}\) ......(4)

 

Now, Assume the total time taken is t.

Hence, using (1), we get,

v=\(\frac{l}{t}\)……(5)

Therefore, by equating (4) and (5), we get,

\(\frac{l}{t}\) =\(l(\frac{t_1t_2}{t_2+t_1})\)

∴t=(\(\frac{t_1 t_2}{t_2+t_1}\))

Hence, the required time taken is  \(\frac{t_1t_2}{t_2+t_1}\)

 

Therefore, the correct answer is (B).