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Chapter 2: Problem 65

At a metro station, a girl walks up a stationary escalator in time \(t_{1}\). If she remains stationary on the escalator, then the escalator take her up in time \(t_{2}\). The time taken by her to walk up on the moving escalator will be [NCERT Exemplar] (a) \(\left(t_{1}+t_{2}\right) / 2\) (b) \(t_{1} t_{2} /\left(t_{2}-t_{1}\right)\) (c) \(t_{1} t_{2} /\left(t_{2}+t_{1}\right)\) (d) \(t_{1}-t_{2}\)

Short Answer

Expert verified
The time taken is \( \frac{t_1 t_2}{t_1 + t_2} \), so the answer is (c).

Step by step solution

01

Identify the Given Times

We are given two scenarios: The girl walks up the stationary escalator in time \( t_1 \) and the escalator, when moving, takes \( t_2 \) to carry someone up without walking.
02

Understand the Problem

The goal is to find the time \( t \) taken by the girl to walk up when the escalator is moving.
03

Define the Relationship Between Speeds

Let the speed of the girl on the escalator be \( v_g \) and the speed of the escalator be \( v_e \). The length of the escalator \( L \) relates to the times and speeds: \( L = v_g t_1 = v_e t_2 \).
04

Find the Combined Speed Equation

When the girl walks on the moving escalator, the effective speed becomes \( v_g + v_e \). The time taken \( t \) will then be the escalator length divided by this combined speed: \( t = \frac{L}{v_g + v_e} \).
05

Substitute the Length and Speeds

Substitute \( L = v_g t_1 \) and \( L = v_e t_2 \) into the combined speed equation: \[ t = \frac{L}{v_g + v_e} = \frac{v_g t_1}{v_g + v_e} = \frac{v_e t_2}{v_g + v_e}. \]
06

Set Expressions for Speeds

From \( L = v_g t_1 \), we have \( v_g = \frac{L}{t_1} \) and from \( L = v_e t_2 \), we get \( v_e = \frac{L}{t_2} \).
07

Simplify the Equation

Substitute \( v_g = \frac{L}{t_1} \) and \( v_e = \frac{L}{t_2} \) into \( t = \frac{L}{v_g + v_e} \):\[ t = \frac{L}{\frac{L}{t_1} + \frac{L}{t_2}} = \frac{L}{\frac{L(t_1 + t_2)}{t_1 t_2}} = \frac{t_1 t_2}{t_1 + t_2}. \]
08

Compare with Options

The expression \( \frac{t_1 t_2}{t_1 + t_2} \) matches option (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed and Velocity
Speed and velocity are fundamental concepts in physics that help us describe and analyze motion. Speed is a scalar quantity, which means it's only concerned with how fast an object is moving, regardless of its direction. It's typically measured in units like meters per second (m/s) or kilometers per hour (km/h).
Velocity, on the other hand, is a vector quantity. This means it provides information about both the speed and direction of an object's movement. Let's say a girl walks up an escalator. Her speed would tell us how many meters she covers per second. Her velocity would also include whether she is moving upwards, downwards, or sideways.
In the exercise, we dealt with both velocity and speed because the girl and the escalator each have respective speeds, which must be combined when considering the overall velocity of the girl on the moving escalator.
Time and Distance
Time and distance play a crucial role in understanding motion, especially when dealing with relative motion. In physics problems like the one given, the concept of time is used to determine how long an action takes, while distance refers to how far an object travels during that time.
In the context of the escalator exercise, we can think of distance as the length of the escalator (denoted as \(L\)). The time it takes for the girl to walk this distance when the escalator is stationary is \(t_1\), and the time the escalator takes to cover the same distance without her walking is \(t_2\).
By combining these times with respective speeds, we can calculate the time the girl needs when the escalator is moving. This involved using the formula \( t = \frac{L}{v_g + v_e} \) where \(v_g\) and \(v_e\) are speeds of the girl and the escalator respectively. This showcases how distance and time integrate to solve motion problems.
Physics Problem Solving
Physics problem solving involves breaking down a complex situation into understandable parts. By utilizing known formulas and concepts, we can determine unknown variables.
Firstly, in the exercise, identifying the given variables and what needs to be solved is crucial. We know the times \(t_1\) and \(t_2\), and need to find the time \(t\) when both speeds work together. By understanding this, a clear path to solving the problem can be formed.
Secondly, forming relationships between the elements in the problem is essential. Knowing that the length of the escalator \(L\) equals the girl鈥檚 speed times \(t_1\) and the escalator鈥檚 speed times \(t_2\) helps in deriving useful equations.
Lastly, substituting and simplifying these equations to find a logical answer, as we did when confirming \(\frac{t_1 t_2}{t_1 + t_2}\) as the solution, is a key step. Physics problem solving not only answers the question but deepens understanding of how motion works.

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Most popular questions from this chapter

The velocity of a particle is \(v=v_{0}+g t+f t^{2} .\) If its position is \(x=0\) at \(t=0\), then its displacement after unit time \((t=1)\) is (a) \(v_{0}-g / 2+f\) (b) \(v_{0}+g / 2+3 f\) (c) \(v_{0}+g / 2+f / 3\) (d) \(v_{0}+g+f\)

A parachutist after alling out falls \(50 \mathrm{~m}\) without friction. When parachute opens, it decelerates at \(2 \mathrm{~ms}^{-2}\). He reaches the ground with a speed of \(3 \mathrm{~ms}^{-1}\). At what height, did he fallen out? [AIEEE 2005] (a) \(111 \mathrm{~m}\) (b) \(293 \mathrm{~m}\) (c) \(182 \mathrm{~m}\) (d) \(91 \mathrm{~m}\)

An object, moving with a speed of \(6.25 \mathrm{~m} / \mathrm{s}\), is declerated at a rate given by \(\frac{d v}{d t}=-2.5 \sqrt{v}\), where \(v\) is the instantaneous speed. The time taken by the object, to come to rest would be (a) \(2 \mathrm{~s}\) (b) \(4 \mathrm{~s}\) (c) \(8 \mathrm{~s}\) (d) \(1 \mathrm{~s}\)

A body freely falling from rest has a velocity \(v\) after it falls through distance \(h .\) The distance it has to fall down further for its velocity to become double is (a) \(h\) (b) \(2 \mathrm{~h}\) (c) \(3 h\) (d) \(4 h\)

Water drops fall from a tap on the floor \(5 \mathrm{~m}\) below at regular intervals of time, the first drop striking the floor when the fifth drop begins to fall. The height at which the third drop will be, from ground, at the instant when first drop strikes the ground, will be \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(1.25 \mathrm{~m}\) (b) \(2.15 \mathrm{~m}\) (c) \(2.73 \mathrm{~m}\) (d) \(3.75 \mathrm{~m}\)
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