/*! 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 24 A car drives along a straight le... [FREE SOLUTION] | 91Ó°ÊÓ

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A car drives along a straight level frictionless road by an engine delivering constant power. Then velocity is directly proportional to \(\begin{array}{ll}\text { (1) } t & \text { (2) } \frac{1}{\sqrt{t}}\end{array}\) (3) \(\sqrt{t}\) (4) None of these

Short Answer

Expert verified
Velocity is directly proportional to \( \sqrt{t} \). Option (3) is correct.

Step by step solution

01

Analyzing Power and Force Relation

The power delivered by the car's engine is constant. Power \( P \) can be expressed as \( P = F imes v \), where \( F \) is the force and \( v \) is the velocity of the car. Because the road is frictionless, the only force considered here is the one provided by the engine.
02

Expressing Force Using Mass and Acceleration

Using Newton's second law, we know that \( F = m imes a \), where \( m \) is the mass of the car and \( a \) is the acceleration. Since \( P = m imes a imes v \), and \( P \) is constant, this equation relates acceleration and velocity under a constant power output.
03

Relating Velocity and Time

Rearrange the equation from Step 2 to express acceleration \( a = \frac{P}{m imes v} \). Acceleration \( a \) is also given by \( a = \frac{dv}{dt} \). Hence, \( \frac{dv}{dt} = \frac{P}{m imes v} \).
04

Solving the Differential Equation

Now, separate variables and integrate: \[ v imes dv = \frac{P}{m} imes dt \]. Integrating both sides, we get \[ \frac{v^2}{2} = \frac{P}{m} imes t + C \] where \( C \) is the integration constant.
05

Solving for Velocity as Function of Time

Solving the equation from Step 4 for velocity, \[ v = \sqrt{\frac{2P}{m} t + C} \]. At \( t = 0 \), if initial velocity \( v_0 = 0 \), then \( C = 0 \). Thus \[ v = \sqrt{\frac{2P}{m} t} = k \sqrt{t} \], where \( k = \sqrt{\frac{2P}{m}} \).
06

Matching with Choices

From Step 5, we concluded that \( v \) is directly proportional to \( \sqrt{t} \). Comparing with the options given, option (3) \( \sqrt{t} \) is correct.

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion is a fundamental principle in physics that describes the relationship between force, mass, and acceleration. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, defined as \( F = m \times a \). This equation tells us how much an object will accelerate when a certain force is applied.

In the context of the given exercise, this law is crucial for understanding how the car accelerates on a frictionless road. Here, the force is supplied by the car's engine. The road being frictionless means the force from the engine is the main factor in driving the car forward.
  • The mass \( m \) is constant, representing the car's mass.
  • The acceleration \( a \) can change, depending on the force applied.

For a car with constant power, changes in velocity are directly tied to how the power affects acceleration, which in turn depends on the velocity itself. This relationship is analyzed further through mathematical equations involving differential equations.
Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. In simple terms, they provide a way to model how changing one variable affects another, especially over time.

In the exercise, we use a differential equation to connect velocity \( v \) with time \( t \) when the power \( P \) is constant. The key relationship here is \( \frac{dv}{dt} = \frac{P}{m \times v} \).
  • \( \frac{dv}{dt} \) represents how velocity changes over time, also known as acceleration.
  • \( \frac{P}{m \times v} \) shows how constant power affects acceleration through velocity.


Solving this differential equation involves integrating both sides of the equation to find \( v \) in terms of \( t \). After integrating, we find that \( v = \sqrt{\frac{2P}{m} t + C} \). By determining the value of the constant \( C \), we can express \( v \) as a clear function of time.
Constant Power
Constant power in this scenario means that the car's engine consistently delivers the same amount of energy over time. Power is a measure of how quickly work is done or energy is transferred, mathematically represented as \( P = F \times v \), where \( F \) is force and \( v \) is velocity.

For a car accelerating on a frictionless surface, constant power affects how its speed changes over time. The crux of the problem is understanding how constant power influences the car's velocity.
  • A constant power output means that as the car speeds up, the force exerted by the engine must decrease, since \( F = \frac{P}{v} \).
  • This creates a dynamic relationship that determines how quickly the car reaches higher speeds.


Because power is constant, the ideal expression for velocity ends up being \( v = k \sqrt{t} \). This indicates that the car's velocity changes in proportion to the square root of time, highlighting the non-linear acceleration typical of constant power scenarios.

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Most popular questions from this chapter

In which of the following cases can the work done increase the potential energy? (1) Both conservative and non-conservative forces (2) Conservative force only (3) Non-conservative force only (4) Neither conservative nor non-conservative forces.

A block of mass \(2 \mathrm{~kg}\) is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force \(F=40\) \(\mathrm{N}\). The kinetic energy of the particle increase \(40 \mathrm{~J}\) in a given interval of time. Then: \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (1) tension in the string is \(40 \mathrm{~N}\) (2) displacement of the block in the given interval of time is \(2 \mathrm{~m}\) (3) work done by gravity is \(-20 \mathrm{~J}\) (4) work done by tension is \(80 \mathrm{~J}\)

The \(\mathrm{PE}\) of a certain spring when stretched from natural length through a distance \(0.3 \mathrm{~m}\) is \(5.6 \mathrm{~J}\). Find the amount of work in joule that must be done on this spring to stretch it through an additional distance \(0.15 \mathrm{~m}\).

A man pushes a block of \(30 \mathrm{~kg}\) along a level floor at a constant speed with a force directed at \(45^{\circ}\) below the horizontal. If the coefficient of friction is \(0.20\), then match the following. $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Work done by all forces exerted by the surface on the block in \(20 \mathrm{~m}\)} & \text { a. } \text{Zero} \\ \hline \text { ii. } \text{Work done by the force of gravity} & \text { b. } \text{\(-1500 \mathrm{~J}\)} \\ \hline \text { iii. } \text{Work done by the man on the block in pushing it through \(10 \mathrm{~m}\)} & \text { c. } \text{\(750 \mathrm{~J}\)}\\\ \hline \text { iv. } \text{Net force on the block} & \text { d. } \text{\(30 \mathrm{~J}\)} \\ \hline \end{array} $$

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 kinetic energy of the particle and the value of \(x\) at which maximum kinetic energy occurs are (1) \(29 \mathrm{~J}, 0 \mathrm{~m}\) (2) \(49 \mathrm{~J}, 0 \mathrm{~m}\) (3) \(49 \mathrm{~J}, 2 \mathrm{~m}\) (4) \(29 \mathrm{~J}, 2 \mathrm{~m}\)

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