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Chapter 3: Problem 100

A person climbs from a Paris metro station to the street level by walking up a stalled escalator in 94 s. It takes \(66 \mathrm{s}\) to ride the same distance when standing on the escalator when it is operating normally. How long would it take for him to climb from the station to the street by walking up the moving escalator?

Short Answer

Expert verified
Answer: It would take approximately 38.58 seconds for the person to climb from the station to the street by walking up the moving escalator.

Step by step solution

01

Assign variables

Let's assign variables for given data: - \(T_w\): the time it takes to walk up the stalled escalator, \(T_w=94\) s - \(T_r\): the time it takes to ride the same distance when the escalator is working, \(T_r=66\) s - \(T_\text{total}\): the time we need to find out for the person to climb from the station to the street by walking up the moving escalator
02

Calculate the distance

Since the distance covered in both cases is the same, we can denote this by the variable \(D\). To find this value, we can use the relationship: \(D = S_w \times T_w\) and \(D = S_e \times T_r\), where \(S_w\) is the person's walking speed and \(S_e\) is the escalator's speed. We can rewrite the equation as: \(S_w = \frac{D}{T_w}\) and \(S_e = \frac{D}{T_r}\)
03

Calculate the relative speed

When the person is walking up the moving escalator, their total speed would be the sum of their walking speed and the escalator's speed because they are moving in the same direction. We can write this as: \(S_\text{total} = S_w + S_e\) Substitute the values for speed from Step 2: \(S_\text{total} = \frac{D}{T_w} + \frac{D}{T_r}\)
04

Calculate the time it takes to climb by walking up the moving escalator

Now we can use the relationship distance = speed × time to find the total time taken by the person while walking up the moving escalator: \(T_\text{total} = \frac{D}{S_\text{total}}\) Substitute the expression for \(S_\text{total}\) from Step 3: \(T_\text{total} = \frac{D}{\frac{D}{T_w} + \frac{D}{T_r}}\) Note that we can cancel the \(D\) variable from the equation: \(T_\text{total} = \frac{1}{\frac{1}{T_w} + \frac{1}{T_r}}\) Plug in the values of \(T_w\) and \(T_r\): \(T_\text{total} = \frac{1}{\frac{1}{94\,\text{s}} + \frac{1}{66\,\text{s}}}\)
05

Calculate the final answer

We can now compute the value for \(T_\text{total}\): \(T_\text{total} = \frac{1}{\frac{94 + 66}{94 \times 66}} = \frac{94 \times 66}{94 + 66} = 38.579\) s Rounded to two decimal places, it would take approximately \(\boldsymbol{38.58}\) seconds for the person to climb from the station to the street by walking up the moving escalator.

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Most popular questions from this chapter

A ball is thrown horizontally off the edge of a cliff with an initial speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) How long does it take for the ball to fall to the ground 20.0 m below? (b) How long would it take for the ball to reach the ground if it were dropped from rest off the cliff edge? (c) How long would it take the ball to fall to the ground if it were thrown at an initial velocity of \(20.0 \mathrm{m} / \mathrm{s}\) but \(18^{\circ}\) below the horizontal?
Show that for a projectile launched at an angle of \(45^{\circ}\) the maximum height of the projectile is one quarter of the range (the distance traveled on flat ground).
A ball is thrown from a point \(1.0 \mathrm{m}\) above the ground. The initial velocity is \(19.6 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. (a) Find the maximum height of the ball above the ground. (b) Calculate the speed of the ball at the highest point in the trajectory.
A vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of \(22.2 \mathrm{cm}\) and makes an angle of \(130.0^{\circ}\) with the positive \(x\) -axis. What are the \(x\) - and \(y\) -components of this vector?
An arrow is shot into the air at an angle of \(60.0^{\circ}\) above the horizontal with a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) What are the \(x\) - and \(y\) -components of the velocity of the arrow \(3.0 \mathrm{s}\) after it leaves the bowstring? (b) What are the \(x\) - and \(y\) -components of the displacement of the arrow during the 3.0 -s interval?
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