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A pcrson travelling on a straight linc moves with a uniform velocity \(v_{1}\) for a distance \(x\) and with a unilorm velocity \(v_{2}\) for the next equal distanee The average velocity \(v\) is given by (a) \(v-\frac{v_{1}+v_{2}}{2}\) (b) \(v-\sqrt{v_{1} v_{2}}\) (c) \(\frac{2}{v}-\frac{1}{v_{1}}+\frac{1}{v_{2}}\) (d) \(\frac{1}{v}-\frac{1}{v_{1}}+\frac{1}{v_{2}}\)

Short Answer

Expert verified
The correct answer is (c) \(\frac{2}{v}-\frac{1}{v_{1}}+\frac{1}{v_{2}}\).

Step by step solution

01

Calculate total distance traveled

Since the person travels a distance \(x\) at velocity \(v_1\) and another distance \(x\) at velocity \(v_2\), the total distance \(d\) is \(d = x + x = 2x\).
02

Calculate time for each section of travel

The time taken to travel the first distance \(x\) with velocity \(v_1\) is \(t_1 = \frac{x}{v_1}\). Similarly, the time taken to travel the second distance \(x\) with velocity \(v_2\) is \(t_2 = \frac{x}{v_2}\).
03

Calculate total time taken

The total time \(t\) for the whole journey is the sum of \(t_1\) and \(t_2\). Therefore, \(t = \frac{x}{v_1} + \frac{x}{v_2}\).
04

Calculate average velocity

The average velocity \(v\) is given by the formula \(v = \frac{\text{total distance}}{\text{total time}}\). Substitute the values from previous steps: \[ v = \frac{2x}{\frac{x}{v_1} + \frac{x}{v_2}} = \frac{2x}{x \left( \frac{1}{v_1} + \frac{1}{v_2} \right)} = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}} \].
05

Simplify to match one of the given choices

From Step 4, we see that \(v\) can be represented as \[ \frac{1}{v} = \frac{1}{2} \left( \frac{1}{v_1} + \frac{1}{v_2} \right) \]. Multiply both sides by 2 to match the form of the options: \[ \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2} \]. Re-arrange to get \( \frac{2}{v} - \frac{1}{v_1} + \frac{1}{v_2} = 0 \), which simplifies to option (c) \( \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2} \).

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Key Concepts

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

Uniform Velocity
Uniform velocity occurs when an object travels equal distances in equal intervals of time without a change in speed or direction. Imagine driving a car at a constant speed on a straight highway without stopping or speeding up. That's uniform velocity! Uniform velocity is important in physics because it allows us to predict and calculate the motion of objects over time.

When dealing with uniform velocity, the formula to find the velocity is simply \[ v = \frac{d}{t} \]where
  • \( v \) is the velocity
  • \( d \) is the distance covered
  • \( t \) is the time taken
For objects moving with uniform velocity, their path can easily be tracked using distance-time graphs, which result in a straight line. This is because equal changes in distance happen over equal intervals of time.

Understanding uniform velocity is crucial for solving various physics problems, such as the one about a person traveling on a straight path with two different velocities.
Distance-Time Relationship
The distance-time relationship is fundamental in physics to describe how an object's position changes over time. This relationship helps us understand motion and calculate other aspects like velocity and time taken for travel.

When we say a person travels a distance of \( x \) at velocity \( v_1 \) and an equal distance at velocity \( v_2 \), we become concerned with two main aspects:
  • Total distance traveled
  • Total time taken for the journey
To break it down:
  • Total distance is simply the sum of the distances covered at each velocity. In this case, it's \( 2x \).
  • To find the total time taken, you add the time spent traveling each section: \( t = \frac{x}{v_1} + \frac{x}{v_2} \).
This is key to solving average velocity problems because it helps us determine how long each part of the journey takes compared to the whole.
JEE Physics Problem
JEE (Joint Entrance Examination) physics problems often test students' understanding of fundamental physics concepts, requiring practical application of theory. This problem, for example, challenges students on their comprehension of average velocity, the concept of uniform velocity, and their ability to apply the distance-time relationship.

To solve this exercise:1. **Understand the Question**: The problem asks for average velocity, given different uniform velocities over equal distances.2. **Use Known Formulas**: Utilize the formula for average velocity: \[ v = \frac{\text{Total Distance}}{\text{Total Time}} \]3. **Calculate Step-by-Step**: Break it down into calculations for total distance and time, combining them to find average velocity.4. **Match to Options**: Simplify the result using algebra to best match the multiple-choice options.
JEE problems like these are not just about getting the right answer—they aim to deepen your understanding of physics and enhance problem-solving skills through practical applications.

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Most popular questions from this chapter

A body is moving with uniform acceleration starting from rest. Its motion is deseribed by some interpreted results. Column-I 1 (p) Column-II \(v_{1}: v_{2} \quad v_{3}=1: 2: 3\) (a) Its velocity at the end \(o^{2} t\) sec., \(2 t\) sec. \(3 t \mathrm{sec}\). are in the ratio (b) Distances covered in \(\mathrm{I}^{* 1} t\) sec., \(2^{\text {all }} t\) sec. (q) \(v_{1}: v_{2} \quad v_{3}=1: 1: 1\) \(3^{\text {ra }} t\) sec. are in the ratio (c) Distances covered in \(t\) sec. \(2 t\) see. (r) \(s_{1}: s_{2}: s_{3}=1: 3: 5\) \(3 t\) sec. are in the ratio (d) Change in velocity at the cnd of \(1^{A}\) (s) \(s_{1}: s_{2}: s_{3}=1: 4: 9\) \(t\) sec. \(2^{\text {nd }} t \sec , 3^{\text {nis }} t\) sec. are in the ratio

Statement-1: Distance-lime graph of the motion of a body having uniformly accelerated motion is a straight line inclined to the time axis. Statement-2 : Distance travelled by a body having uniformly accelerated motion is directly proportional to the square of the time taken.

A body is projected vertically upwards from the ground. Taking air resistanec into account: (a) the time of ascent and maximum height is less than that in vaccuum. (b) the time of ascent is greater than the time of descent. (c) the time of ascent is less than the time of descent. (d) the speed of the particle when it returns to the ground is less than the speed of projection.

Statement-1 : The acceleration of a body is always zero, where instantaneous velocity is zero. Statement-2 : The acceleration is a scalar quantity.

\(\Lambda\) stone is dropped from a balloon going up with a uniform velocity of \(5.0 \mathrm{~m} / \mathrm{s}\). If the balloon was \(50 \mathrm{~m}\) high when the stone was dropped, find its height when the stone hits the ground. 'lake \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\).

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