Question

Naina walks across a stationary escalator in 5 seconds. If she stands on a moving escalator, it
takes her 2 seconds to reach the other end. If Naina walks on a moving escalator, how long
does it take her to cross over to the other end?

Answer 1

Answer:

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Explanation:

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Answer 2

It takes 1.43 seconds for Naina to cross the moving escalator while walking on it.

Given,

Time in which Naina walks across the stationary accelerator=5 seconds

Time in which she reaches the other end while standing on the moving accelerator=2 seconds.

To find,

the time taken to cross when she walks on a moving accelerator.

Solution:

• The motion between the moving escalator and Naina walking on it is a downstream motion.
• In a downstream motion, the resultant speed of the motion is equal to the sum of the individual speeds of the components present in the motion.
• Vnet=V1+V2.

First case: When she walks on the stationary escalator.

Speed of escalator=0(stationary).

Naina's speed=Vn.

Let the length of the escalator(the distance covered by Naina) be D.

$Speed=\frac{Distance}{Time} \\Vn=\frac{D}{5}\\D=VnX5\\D=5Vn m.$

As the escalator is same throughout, D will also be same.

Second case: When she stands on the moving escalator.

Speed of escalator=Ve.

Naina's speed=0(she just stands).

$Speed=\frac{Distance}{Time} \\Ve=\frac{D}{2}\\D=VeX2\\D=2Ve m.$

Third case: When she walks on the moving accelerator.

The resultant speed, Vnet will be:

Vnet=Vn+Ve.

D will still be the same so,

$Speed=\frac{Distance}{Time} \\Distance=SpeedXTime\\D=VnetXTime\\D=(Vn+Ve)XTime\\D=(\frac{D}{5} +\frac{D}{2} )XTime\\1=(\frac{1}{5} +\frac{1}{2} )XTime\\1=(\frac{2}{10} +\frac{5}{10} )XTime\\1=\frac{7}{10}XTime\\Time=\frac{10X1}{7} \\Time=\frac{10}{7}\\Time=1.43 seconds.$

Thus, it takes 1.43 seconds.

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