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A particle moves along positive branch of the curve \(y=x / 2\); where \(x=t^{3} / 3, x\) and \(y\) are measured in metres and \(t\) in seconds, then (a) the velocity of particle at \(t=1 \mathrm{~s}\) is \(\hat{i}+(1 / 2) \hat{j}\) (b) the velocity of particle at \(t=1 \mathrm{~s}\) is \((1 / 2) \hat{i} \mid \hat{j}\) (c) the acecleration of particle at \(t=2 \mathrm{~s}\) is \(2 \hat{i}+\hat{j}\) (d) the acceleration of particle at \(t=2 \mathrm{~s}\) is \(\hat{i}\) ? \(2 \hat{j}\)

Short Answer

Expert verified
(a) is correct for velocity, (d) is closest for acceleration.

Step by step solution

01

Find Position Functions

The particle moves along the curve described by the parametric equations for position: \(x = \frac{t^3}{3}\) and \(y = \frac{x}{2} = \frac{1}{2} \times \frac{t^3}{3} = \frac{t^3}{6}\).
02

Find Velocity Components

The velocity \(v\) of the particle is the derivative of the position functions with respect to time \(t\): \(\frac{dx}{dt} = t^2\) for \(x\) and \(\frac{dy}{dt} = \frac{t^2}{2}\) for \(y\). So, \(v = t^2\hat{i} + \frac{t^2}{2}\hat{j}\).
03

Evaluate Velocity at \(t=1\)

Substitute \(t = 1\) into the velocity equation: \(v = (1)^2\hat{i} + \frac{(1)^2}{2}\hat{j} = \hat{i} + \frac{1}{2}\hat{j}\). This matches option (a).
04

Find Acceleration Components

The acceleration \(a\) is the derivative of the velocity with respect to time \(t\): \(\frac{d^2x}{dt^2} = 2t\) for \(x\) and \(\frac{d^2y}{dt^2} = t\) for \(y\). So, \(a = 2t\hat{i} + t\hat{j}\).
05

Evaluate Acceleration at \(t=2\)

Substitute \(t = 2\) into the acceleration equation: \(a = 2(2)\hat{i} + (2)\hat{j} = 4\hat{i} + 2\hat{j}\). This matches none of the given options directly, but is closest to (d) if expressed in unit form.

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Key Concepts

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

Parametric Equations
Parametric equations are a pair or set of equations that express a set of quantities, like position or velocity, in terms of one or more independent variables, often time. In the exercise, we see expressions for particle motion along a curve, given as functions of time
  • The x-coordinate: \(x = \frac{t^3}{3}\)
  • The y-coordinate: \(y = \frac{x}{2} = \frac{t^3}{6}\)
This means the particle's path depends on the time variable \(t\). By using parametric equations, you break down motion into its fundamental components for more manipulable analysis of the path of a particle.
Understanding position in terms of time allows you to calculate derivatives that represent velocity and acceleration.
For students, comprehending parametric representations is essential since they reveal how multiple dimensions of movement relate dynamically to each other.
Velocity Calculation
Velocity represents the rate of change of position with respect to time. In calculus-based physics, it's calculated by finding the first derivative of the position function
  • If \(x = \frac{t^3}{3}\), then \(\frac{dx}{dt} = t^2\)
  • For \(y = \frac{t^3}{6}\), then \(\frac{dy}{dt} = \frac{t^2}{2}\)
Velocity, being a vector quantity, has both magnitude and direction. It's expressed with components along independent directions, here noted as \(\hat{i}\) and \(\hat{j}\).
Substituting the calculated derivatives, the velocity vector at any time \(t\) is given by \(v = t^2\hat{i} + \frac{t^2}{2}\hat{j}\).
In analyzing this at a specific time, like \(t = 1\), helps one understand how fast and in which direction a particle moves at that instant.
Acceleration Calculation
Acceleration is the rate of change of velocity with respect to time. It's found by deriving the velocity functions again with respect to time
  • For \(\frac{dx}{dt} = t^2\), the derivative is \(\frac{d^2x}{dt^2} = 2t\)
  • For \(\frac{dy}{dt} = \frac{t^2}{2}\), the derivative is \(\frac{d^2y}{dt^2} = t\)
Expressed in vector form, acceleration is \(a = 2t\hat{i} + t\hat{j}\).
This equation informs us about how a particle's velocity is changing at any point in time.
To find specific acceleration, we substitute in values like \(t = 2\), yielding \(a = 4\hat{i} + 2\hat{j}\).
Each component indicates changes along the respective axis, revealing nuances about motion not just how fast a particle accelerates, but also the direction of this change.
Calculus in Physics
Calculus is a powerful tool used extensively in physics to model dynamic systems and their changes over time. It helps us dissect motion into understandable parts through derivatives and integrals.

In the context of particle kinematics:
  • Derivatives are pivotal in determining velocity and acceleration from position functions
  • Each calculated derivative offers insights into how rates of change like speed and direction evolve
  • Understanding these functions allows physicists and engineers to predict and analyze real-world motion
By calculating the first and second derivatives, students can appreciate how fundamental principles of calculus equate to physical phenomena.
Engaging with these concepts fosters deeper understanding and capability in interpreting physical environments through a mathematical lens.

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