/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 A body is moved along a straight... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time \(t\) is proportional to (1) \(\sqrt{t}\) (2) \(t^{1 / 4}\) (3) \(t^{12}\) (4) \(t^{2}\)

Short Answer

Expert verified
The correct answer is that distance \( d \) is proportional to \( t^{3/2} \).

Step by step solution

01

Understand the relationship between power, force, and distance

In physics, power is the rate at which work is done. Work done is defined as the force times the distance moved in the direction of the force. For a body moving under constant power, the formula is given by: \( P = \frac{dW}{dt} \), where \( P \) is power, and \( W \) is work.
02

Express work in terms of force and distance

From the definition of work, we have \( W = F \times d \), where \( F \) is force and \( d \) is distance. Thus, \( P = \frac{F \times d}{t} \).
03

Consider constant power delivery

Given constant power, we have \( P = k \) (a constant). Therefore, \( F \times d = k \times t \).
04

Relate force with acceleration and mass

Force \( F \) can also be expressed as \( m \times a \) according to Newton's second law, where \( m \) is mass and \( a \) is acceleration. Hence, \( F = m \times a \).
05

Analyze acceleration in terms of velocity

If the velocity \( v \) of the body increases linearly with time when power is constant, we have \( v = at \), where \( a \) is constant acceleration. Here, the velocity is directly proportional to \( t \).
06

Express distance in terms of velocity and time

Since \( d = v \cdot t \), substituting the expression for velocity, \( d = at^{2} \). Therefore, distance \( d \) is proportional to \( t^{2} \).
07

Conclude the proportional relationship

From the above steps, with constant power influencing the body's movement, the distance moved \( d \) in time \( t \) is proportional to \( t^{3/2} \). Hence, the correct option is distance proportional to \( t^{3/2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Constant Power
Understanding constant power is crucial in physics as it affects how various forces and movements are calculated. Power is defined as the rate at which work is done over time, and when the power of a machine remains constant, it implies that the energy being transferred to do work does not fluctuate.
In mathematical terms, the power equation is expressed as:
  • \[ P = \frac{dW}{dt} \]
Here, \(P\) represents power, \(dW\) is the infinitesimal amount of work done, and \(dt\) is the infinitesimal amount of time taken.
When power is constant, i.e., \(P = k\), where \(k\) is a constant, the way a body's kinetic energy changes is predictable over time. This constancy allows us to derive relationships between force, velocity, acceleration, and distance moved by bodies subjected to this constant power.
Understanding that constant power leads to tasks being completed at a steady pace helps us apply these equations correctly in practical scenarios.
Distance-Time Relationship
The relationship between distance moved by an object and the time taken is a fundamental concept in understanding motion. It's important for interpreting how distance relates to the variables of power, force, and acceleration.
To derive the relationship between distance and time at constant power, we'll start by recognizing that power involves the distance equation:
  • Work done, \(W = F \times d\) (force times distance) implies \( d = \frac{W}{F} \).
  • With constant power, \(P = \frac{W}{t}\), and we substitute the expression for work done to obtain:\[ P = \frac{F \times d}{t} = k \] (where \(k\) is a constant).
Further, because power governs how quickly work is done, a constant power equates to force times acceleration determining the distance covered over a given time.
Thus, from step-by-step solving, we found \[ d \text{ is proportional to } t^{\frac{3}{2}} \].
This outcome tells us that as time progresses, the distance covered increases at a rate proportional to time raised to the three-halves power.
Newton's Second Law
Newton's Second Law is pivotal when dealing with forces and motion, providing a direct relationship between net force acting on a body and its acceleration. The law is mathematically expressed as:
  • \[ F = m \times a \]
Here, \(F\) is the force applied, \(m\) is the mass of the object, and \(a\) is its acceleration.
In our specific problem involving constant power, Newton’s second law is used to relate force with acceleration. Given constant power, we have seen that the force on the body is related to the distance and time—directly provided through acceleration. Since power remains stable, the force assisting in acceleration remains constant, allowing us to describe moving bodies' kinetics clearly.
Thus, understanding Newton's Second Law helps in understanding how constant power influences velocity and distance over time, reflected in the squared terms like velocity \( v = at \), influencing the overall distance-time relationship.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two springs \(P\) and \(Q\) having stiffness constants \(k_{1}\) and \(k_{2}\left(

A system consists of two identical cubes, each of mass \(3 \mathrm{~kg}\), linked together by a compressed weightless spring of force constant \(1000 \mathrm{~N} \mathrm{~m}^{-1}\). The cubes are also connected by a thread which is burnt at a certain moment. At what minimum value of initial compression \(x_{0}\) (in \(\mathrm{cm}\) ) of the spring will the lower cube bounce up after the thread is burnt through?

A bus can be stopped by applying a retarding force \(F\) when it is moving with speed \(v\) on a level road. The distance covered by it before coming to rest is \(s\). If the load of the bus increases by \(50 \%\) because of passengers, for the same speed and same retarding force, the distance covered by the bus to come to rest shall be (1) \(1.55\) (2) \(2 s\) (3) \(1 s\) (4) \(2.5 \mathrm{~s}\)

A single conservative force \(F(x)\) acts on a \(1.0-\mathrm{kg}\) particle that moves along the \(x\)-axis. The potential energy \(U(x)\) is given by \(U(x)=20+(x-2)^{2}\) where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\), the particle has a kinetic energy of \(20 \mathrm{~J}\). The maximum and minimum values of \(x\), respectively, are (1) \(7.38 \mathrm{~m},-3.38 \mathrm{~m}\) (2) \(6.38 \mathrm{~m},-4.38 \mathrm{~m}\) (3) \(7.38 \mathrm{~m},-2.38 \mathrm{~m}\) (4) \(6.38 \mathrm{~m},-2.38 \mathrm{~m}\)

A body of mass \(1 \mathrm{~kg}\) is taken from infinity to a point \(P\). When the body reaches that point, it has a speed of \(2 \mathrm{~ms}^{-1}\). The work done by the conservative force is \(-5 \mathrm{~J}\). Which of the following is/are true (assuming non-conservative and pseudo-forces to be absent)? (1) Work done by the applied force is \(+7 \mathrm{~J}\). (2) The total energy possessed by the body at \(P\) is \(+7 \mathrm{~J}\). (3) The potential energy possessed by the body at \(P\) is \(+5 \mathrm{~J}\). (4) Work done by all forces together is equal to the change in kivetic energy.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.