/*! 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 55 Acceleration of a particle, star... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 55

Acceleration of a particle, starting from rest in straight line, changes with time as \(a=6 t \mathrm{~m} / \mathrm{s}^{2}\). Displacement of the particle at \(t=2 \mathrm{~s}\), will be (A) \(24 \mathrm{~m}\) (B) \(8 \mathrm{~m}\) (C) \(16 \mathrm{~m}\) (D) \(4 \mathrm{~m}\)

Short Answer

Expert verified
The displacement of the particle at \(t = 2 \: s\) is \(8 \: \mathrm{m}\). The correct answer is (B) \(8 \mathrm{~m}\).

Step by step solution

01

Integrate the acceleration function to find the velocity function

Since the acceleration function is given by \(a(t) = 6t\), we need to integrate it with respect to time to find the velocity function, \(v(t)\). \[v(t) = \int a(t) dt = \int 6t dt\] Integrating the function, we get: \[v(t) = 3t^2 + C_1\] Since the particle starts from rest, its initial velocity is zero (\(v(0) = 0\)). We can use this information to find the constant of integration, \(C_1\). \[0 = 3(0)^2 + C_1 \Rightarrow C_1 = 0\] So, the velocity function is: \[v(t) = 3t^2\]
02

Integrate the velocity function to find the displacement function

Now that we have the velocity function, we need to integrate it with respect to time to find the displacement function, \(s(t)\). \[s(t) = \int v(t) dt = \int 3t^2 dt\] Integrating the function, we get: \[s(t) = t^3 + C_2\] Since the particle starts from the origin, its initial displacement is zero (\(s(0) = 0\)). We can use this information to find the constant of integration, \(C_2\). \[0 = (0)^3 + C_2 \Rightarrow C_2 = 0\] Therefore, the displacement function is: \[s(t) = t^3\]
03

Find the displacement at \(t = 2 \: s\)

To find the displacement at \(t = 2 \: s\), we evaluate the displacement function at \(t = 2\). \[s(2) = (2)^3 = 8 \: \mathrm{m}\] So, the displacement of the particle at \(t = 2 \: s\) is \(8 \: \mathrm{m}\). The correct answer is (B) \(8 \mathrm{~m}\).

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 91影视!

Key Concepts

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

Acceleration
When we talk about acceleration in kinematics, we are referring to how fast the velocity of an object changes with time. Here, the acceleration is given by the equation \( a(t) = 6t \), which means that the acceleration increases linearly with time. The unit for acceleration is meters per second squared (m/s虏). This linear relation signifies that the longer the particle has been in motion, the stronger its acceleration becomes.

Understanding acceleration is crucial because it is the starting point for determining how other kinematic variables, like velocity and displacement, will behave over time. In this specific problem, the acceleration is a function of time, which is not constant. Thus, by integrating this function, we find the velocity function, establishing how velocity evolves as time progresses.
  • Acceleration is the rate of change of velocity over time.
  • Given function: \( a(t) = 6t \).
  • The higher the time, the greater the acceleration due to the direct relationship.
Velocity function
Velocity is a vector quantity that refers to the rate at which an object changes its position. In this problem, to find the velocity function, we need to integrate the acceleration function \( a(t) = 6t \) with respect to time. This process gives us the velocity function, \( v(t) = 3t^2 \), after solving the integral and recognizing the initial condition.

Importantly, the initial velocity of the particle is zero because it starts from rest. This is why the constant of integration \( C_1 \) is zero after checking our initial condition, \( v(0) = 0 \).

Understanding the velocity function allows us to predict the speed and direction of the particle at any given time. It's a crucial piece in solving for other kinematic elements, such as displacement.
  • Velocity indicates how fast displacement changes over time.
  • Derived function: \( v(t) = 3t^2 \).
  • The initial velocity condition is used to solve for integration constants.
Displacement calculation
Displacement is the change in position of the particle along the line of motion. To find the displacement as a function of time, we integrate the velocity function, \( v(t) = 3t^2 \). This yields the displacement function, \( s(t) = t^3 \). Here, the constant \( C_2 \) is zero because we assume the particle starts at the origin, leading to \( s(0) = 0 \).

Calculating the displacement at a specific time, such as \( t = 2 \text{ s} \), involves substituting this time value into the displacement function: \( s(2) = (2)^3 = 8 \) meters. Thus, the particle has moved 8 meters from its starting point after 2 seconds. Remember, displacement is different from distance as it considers only the initial and final positions, disregarding the path taken.
  • Displacement is the integral of the velocity function.
  • Important for understanding how far an object has moved over time.
  • The calculated displacement at \( t = 2 \text{ s} \) is \( 8 \text{ m} \).

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

The maximum height of a projectile for two complementary angles of projection is \(50 \mathrm{~m}\) and \(30 \mathrm{~m}\) respectively. The initial speed of projectile is (A) \(10 \sqrt{34} \mathrm{~m} / \mathrm{s}\) (B) \(40 \mathrm{~m} / \mathrm{s}\) (C) \(20 \mathrm{~m} / \mathrm{s}\) (D) \(10 \mathrm{~m} / \mathrm{s}\)

A swimmer wishes to cross a \(800 \mathrm{~m}\) wide river flowing at \(6 \mathrm{~km} / \mathrm{hr}\). His speed with respect to water is \(4 \mathrm{~km} / \mathrm{hr}\). He crosses the river in shortest possible time. He is drifted downstream on reaching the other bank by a distance of (A) \(800 \mathrm{~m}\) (B) \(1200 \mathrm{~m}\) (C) \(400 \sqrt{13} \mathrm{~m}\) (D) \(2000 \mathrm{~m}\)

A projectile is fired with a velocity \(u\) at right angle to a slope, which is inclined at an angle \(\theta\) with the horizontal. The range of the projectile on the incline is (A) \(\frac{2 u^{2} \sin \theta}{g}\) (B) \(\frac{2 u^{2}}{g} \tan \theta \sec \theta\) (C) \(\frac{u^{2}}{g} \sin 2 \theta\) (D) \(\frac{2 u^{2}}{g} \tan \theta\)

The initial velocity of a body moving along a straight line is \(7 \mathrm{~m} / \mathrm{s}\). It has a uniform acceleration of \(4 \mathrm{~m} / \mathrm{s}^{2}\). The distance covered by the body in the 5 th second of its motion is (A) \(25 \mathrm{~m}\) (B) \(35 \mathrm{~m}\) (C) \(50 \mathrm{~m}\) (D) \(85 \mathrm{~m}\)

A particle moving in a straight line has velocity and displacement equation as \(v=4 \sqrt{1+s}\), where \(v\) is in \(\mathrm{m} / \mathrm{s}\) and \(s\) is in \(\mathrm{m}\). The initial velocity of the particle is (A) \(4 \mathrm{~m} / \mathrm{s}\) (B) \(16 \mathrm{~m} / \mathrm{s}\) (C) \(2 \mathrm{~m} / \mathrm{s}\) (D) Zero
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Medical Physics

Read Explanation

Particle Model of Matter

Read Explanation

Modern Physics

Read Explanation

Linear Momentum

Read Explanation

Fields in Physics

Read Explanation
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.