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Chapter 3: Problem 31

A particle starts from rest with acceleration \(2 \mathrm{~m} / \mathrm{s}^{2}\). The acceleration of the particle decreases down to zero uniformly during time-interval of 4 second. The velocity of particle after 2 second is : (a) \(3 \mathrm{~m} / \mathrm{s}\) (b) \(4 \mathrm{~m} / \mathrm{s}\) (c) zero (d) \(8 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The velocity of the particle after 2 seconds is 3 m/s.

Step by step solution

01

Understanding the Problem

We need to find the velocity of a particle after 2 seconds given that it starts from rest with an initial acceleration, which decreases linearly to zero over 4 seconds. The task is to calculate the velocity at the 2-second mark.
02

Determine the Formula for Acceleration

The particle starts from rest with an initial acceleration of 2 m/s虏, and over 4 seconds, the acceleration decreases linearly to 0. The linear decrease can be expressed by the formula: \[ a(t) = 2 - \left(\frac{2}{4}\right)t = 2 - 0.5t \]This gives the acceleration at any time \( t \).
03

Integrate the Acceleration Function

To find the velocity, we need to integrate the acceleration over time. The velocity \( V(t) \) as a function of time \( t \) is given by the integral of the acceleration:\[ V(t) = \int (2 - 0.5t) \, dt = \left[ 2t - 0.5 \frac{t^2}{2} \right] = 2t - 0.25t^2 + C \]Since the particle starts from rest, \( V(0) = 0 \), so \( C = 0 \). Thus, the velocity function becomes:\[ V(t) = 2t - 0.25t^2 \]
04

Calculate Velocity at 2 Seconds

Substitute \( t = 2 \) into the velocity equation to find the velocity at that time:\[ V(2) = 2(2) - 0.25(2)^2 = 4 - 1 = 3 \, \text{m/s} \]

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Key Concepts

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

Understanding Acceleration
Acceleration is a measure of how quickly the velocity of an object changes over time. It plays a key role in understanding the motion of objects. When an object speeds up, slows down, or changes direction, it is experiencing acceleration.
The formula for acceleration in this problem is a linear function:
  • Initial acceleration: 2 m/s虏
  • Decreases uniformly to zero over 4 seconds
This tells us that the acceleration at any given time, denoted by \(a(t)\), can be calculated using the equation: \[a(t) = 2 - 0.5t\] Here, \(t\) represents time in seconds. This formula ensures that at \(t=0\), the acceleration is 2 m/s虏, and at \(t=4\), the acceleration becomes zero. Understanding this formula helps us predict how the particle's speed changes over time.
Delving into Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing the motion. It is foundational in solving problems involving velocity and acceleration.
In this context, kinematics helps us bridge the gap between acceleration and velocity. Since velocity is the accumulation (or integral) of acceleration over time, we need to apply principles from kinematics to calculate the velocity at any given point in time.
The velocity function derived in kinematics accounts for initial conditions, such as starting from rest. For this particle, the velocity \(V(t)\) is determined through integration: \[V(t) = \int (2 - 0.5t) \, dt = 2t - 0.25t^2\] This formula allows us to determine the particle's velocity at any moment by substituting the desired time.
Integral Calculus and Its Application
Integral calculus is a crucial mathematical tool in physics, especially in calculating quantities like velocity from acceleration. In our problem, knowing the acceleration at different times isn't enough 鈥� we want to know the velocity, which we derive by integrating the acceleration function.
The velocity function \(V(t) = 2t - 0.25t^2\) arose from performing integration. The process undergoes these steps:
  • Integrate the acceleration function: \(\int (2 - 0.5t) \, dt\)
  • This yields: \(2t - 0.25t^2 + C\)
  • Apply initial conditions: Since the particle starts from rest, \(V(0) = 0\) makes \(C = 0\)
This integration process is fundamental to solving many physics problems involving rates of change. It transforms the given acceleration into the practical velocity equation used to find the velocity at any point, such as after 2 seconds in our exercise.

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Most popular questions from this chapter

During the shooting of a super hit film 'MARD' Amitabh Bachchan was waiting for his beloved 'Amrita Singh' with his dog. When he saw her approaching, the dog was excited and dashed to her then back to master and so on, never stopping. How far would you estimate the dog ran if its speed is \(30 \mathrm{~km} / \mathrm{hr}\) and each of them walked at \(4 \mathrm{~km} / \mathrm{hr}\), starting from a distance \(400 \mathrm{~m}\) apart? (a) \(400 \mathrm{~m}\) (b) \(880 \mathrm{~m}\) (c) \(1500 \mathrm{~m}\) (d) \(30 \mathrm{~km}\)

A particle moves along the positive branch of the curve \(y=\frac{x^{2}}{2}\) where \(x=\frac{t^{2}}{2}\), where \(x\) and \(y\) are measured in metre and \(t\) in second. At \(t=2 \sec\), the velocity of the particle is: (a) \((2 \hat{1}-4 \hat{j}) \mathrm{m} / \mathrm{sec}\) (b) \(\left.\left(2 \hat{1}+4^{2}\right)\right) \mathrm{m} / \mathrm{sec}\) (c) \((2 \hat{1}+2 \hat{j}) \mathrm{m} / \mathrm{sec}\) (d) \((4 \hat{1}-2 \hat{j}) \mathrm{m} / \mathrm{sec}\)

A particle is projected at an angle \(\alpha\) with the horizontal from the foot of an inclined plane making an angle \(\beta\) with horizontal. Which of the following expression holds good if the particle strikes the inclined plane normally? (a) \(\cot \beta=\tan (\alpha-\beta)\) (b) \(\cot \beta=2 \tan (\alpha-\beta)\) (c) \(\cot \alpha=\tan (\alpha-\beta)\) (d) \(\cot \alpha=2 \tan (\alpha-\beta)\)

Two cars move in the same direction along parallel roads. One of them is a \(200 \mathrm{~m}\) long travelling with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). The second one is \(800 \mathrm{~m}\) long travelling with a velocity of \(7.5 \mathrm{~m} / \mathrm{s}\). How long will it take for the first car to overtake the second car? (a) \(20 \mathrm{~s}\) (b) \(40 \mathrm{~s}\) (c) \(60 \mathrm{~s}\) (d) \(80 \mathrm{~s}\)

A particle is projected with speed \(v\) at an angle \(\theta\) \(\left(0<\theta<\frac{\pi}{2}\right)\) above the horizontal from a height \(H\) above the ground. If \(v=\) speed with which particle hits the ground and \(t=\) time taken by particle to reach ground, then: (a) as \(\theta\) increases, \(v\) decreases and \(t\) increases (b) as \(\theta\) increases, \(v\) increases and \(t\) increases (c) as \(\theta\) increases, \(v\) remains same and \(t\) increases (d) as \(\theta\) increases, \(v\) remains same and \(t\) decreases
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