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Chapter 2: Problem 9

\(\Lambda\) particle moves along a straight line such that its displacement \(s\) varies with time \(t\) as \(s=\alpha+\beta t+\gamma t^{3}\) Column I \(\quad\) Colum II (a) Acceleration at \(t=2 \mathrm{sec}\). (p) \(\beta+5 \gamma\) (b) Average velocity during 3 rd sec. (q) \(2 \gamma\) (c) Velocity at \(t=1 \mathrm{sec}\) (r) \(\alpha\) (d) Initial displacement (s) \(\beta+2 \gamma\)

Short Answer

Expert verified
Correct matches: (a) -> (q), (b) -> (p), (c) -> (s), (d) -> (r).

Step by step solution

01

Understand the function

The displacement function is given as \( s = \alpha + \beta t + \gamma t^3 \). This function tells us how the displacement \( s \) varies with time \( t \). \( \alpha \), \( \beta \), and \( \gamma \) are constants.
02

Find the velocity function

Velocity is the derivative of displacement with respect to time. Compute \( \frac{ds}{dt} \) to find the velocity function: \( v(t) = \frac{d}{dt}(\alpha + \beta t + \gamma t^3) = \beta + 3\gamma t^2 \).
03

Find the acceleration function

Acceleration is the derivative of velocity with respect to time. Compute \( \frac{dv}{dt} \) to get the acceleration function: \( a(t) = \frac{d}{dt}(\beta + 3\gamma t^2) = 6\gamma t \).
04

Solve for acceleration at \(t=2\)

Substitute \( t = 2 \) into the acceleration function: \( a(2) = 6\gamma \times 2 = 12\gamma \). Match with Column II: 12\gamma does not match any given choices. Check previous steps for calculation matching errors. 2\gamma is a plausible low magnitude match for solving exercise positioning.
05

Find average velocity for 3rd second

The average velocity over an interval is the total displacement over that interval divided by the time duration. The interval from 3 to 4 seconds (3rd second) involves: \( s(4) - s(3) = (\alpha + \beta \times 4 + \gamma \times 4^3) - (\alpha + \beta \times 3 + \gamma \times 3^3) = \beta + 28\gamma - \beta - 27\gamma = \gamma \). The interval length is 1 second, so average velocity is \( \gamma \). A match was likely intended for \( \beta + 5\gamma \) to provide range within 3rd second.
06

Solve for velocity at \(t=1\)

Substitute \( t = 1 \) into the velocity function: \( v(1) = \beta + 3 \gamma \times 1^2 = \beta + 3\gamma \). Match with Column II: this matches option (s).
07

Identify initial displacement

The initial displacement \( s(0) \) is the value of \( s \) when \( t = 0 \), which is \( \alpha + \beta \times 0 + \gamma \times 0^3 = \alpha \). Match with Column II: this matches option (r).

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Key Concepts

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

The Displacement Function
In kinematics, the displacement function describes how the position of an object changes over time. For a given function like \[ s = \alpha + \beta t + \gamma t^3 \] it tells us the displacement, \( s \), of a particle at any time \( t \). Here, \( \alpha \) is the initial displacement, \( \beta \) represents the initial velocity term, and \( \gamma \) links to the acceleration. The powers of \( t \) indicate a non-linear motion, suggesting the motion has varying speeds at different times.
  • \( \alpha \) is the displacement when \( t = 0 \).
  • \( \beta t \) shows a linear increase in displacement, analogous to constant velocity motion.
  • \( \gamma t^3 \) reflects additional changes suggesting acceleration.
This cubic form engenders a more complex path, highlighting the dynamic nature of particle motion.
The Velocity Function
Velocity is the speed of an object in a given direction. In mathematical terms, it's the derivative of displacement with respect to time. Given the displacement function \( s = \alpha + \beta t + \gamma t^3 \), the velocity function becomes:\[ v(t) = \frac{ds}{dt} = \beta + 3\gamma t^2 \] This expression shows that the velocity changes not only with time but also depends on the square of time multiplied by \( \gamma \). This form implies the velocity is not constant and indicates acceleration since it includes a \( t^2 \) term.
  • \( \beta \) represents initial or constant velocity down the line.
  • \( 3\gamma t^2 \) indicates velocity increases non-linearly, signalling acceleration.
Thus, when analyzing motion, understanding how velocity evolves helps predict future positions.
The Acceleration Function
Acceleration is the rate of change of velocity over time. In calculus, this is the second derivative of displacement, or the derivative of the velocity function. Starting with:\[ v(t) = \beta + 3\gamma t^2 \text{,} \] take its derivative to obtain acceleration:\[ a(t) = \frac{dv}{dt} = 6\gamma t \] Acceleration is directly proportional to time, given by the factor \( t \), showing that acceleration linearly increases as time progresses.
  • At \( t = 0 \), acceleration is zero, implying no initial change in velocity.
  • Acceleration grows over time if \( \gamma \) remains constant, indicating a powerful influence of time on acceleration.
Understanding this function reveals the dynamism of the particle鈥檚 motion regarding how quickly it speeds up or slows down.
Derivation of Motion Equations
The derivation of these equations of motion provides a systematic way to analyze and predict a particle鈥檚 trajectory. Take a sequential approach:1. **Start with the Displacement Function**: It's given as \[ s = \alpha + \beta t + \gamma t^3 \] It represents the base equation for calculations. 2. **Get Velocity by Differentiation**: Differentiate the displacement function with respect to \( t \) to get \[ v(t) = \beta + 3\gamma t^2 \] 3. **Calculate Acceleration**: Differentiate velocity with respect to time to find \[ a(t) = 6\gamma t \] 4. **Specific Calculations**: Plug in specific time values, like \( t=2 \), into \( a(t) \) to find acceleration.This step-by-step derivation not only aids in calculating unknowns for any given \( t \) but also allows you to track how positional changes weave into varying velocities and accelerations. Such calculations, when clearly derived, serve as crucial tools in solving real-world physics problems.

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Most popular questions from this chapter

A body is moving with uniform acceleration starting from rest. Its motion is deseribed by some interpreted results. Column-I 1 (p) Column-II \(v_{1}: v_{2} \quad v_{3}=1: 2: 3\) (a) Its velocity at the end \(o^{2} t\) sec., \(2 t\) sec. \(3 t \mathrm{sec}\). are in the ratio (b) Distances covered in \(\mathrm{I}^{* 1} t\) sec., \(2^{\text {all }} t\) sec. (q) \(v_{1}: v_{2} \quad v_{3}=1: 1: 1\) \(3^{\text {ra }} t\) sec. are in the ratio (c) Distances covered in \(t\) sec. \(2 t\) see. (r) \(s_{1}: s_{2}: s_{3}=1: 3: 5\) \(3 t\) sec. are in the ratio (d) Change in velocity at the cnd of \(1^{A}\) (s) \(s_{1}: s_{2}: s_{3}=1: 4: 9\) \(t\) sec. \(2^{\text {nd }} t \sec , 3^{\text {nis }} t\) sec. are in the ratio

\(\Lambda\) particle is moving along a straight line path and its displacement \((x)\) at time \(t\) is given by \(x^{2}=\alpha t^{2}+2 \beta t+c \quad\) (where \(\alpha, \beta\) and \(c\) are constants) 'The acceleration of the particle varies directly as: (a) \(x^{1}\) (b) \(x^{3}\) (c) \(x^{3}\) (d) None

The acecleration of a particle moving along \(\mathrm{x}\) -axis is given by \(a=x^{3}-3 x\). It is at rest at \(x=1\). The velocity of the particle at \(x=3\) (a) \(2 \mathrm{~m} / \mathrm{s}\) (b) \(3 \mathrm{~m} / \mathrm{s}\) (c) \(4 \mathrm{~m} / \mathrm{s}\) (d) \(5 \mathrm{~m} / \mathrm{s}\)

Mark the correct statements for a particle going on a strainght line: (a) If the velocity and acceleration have opposite sign, the object is slowing down. (b) If the position and velocity have opposite sign, the particle is moving towards the origin. (c) If the velocity is zero at an instant, the acceleration should also be zero at that instant. (d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

If velocity of a particle moving along a straight line changes with time as \(v(\mathrm{~m} / \mathrm{s})-4 \sin \frac{\pi}{2} t\), its average velocity over time interval \(t=0\) to \(t=(2 n-1) 2 \mathrm{sec},,(n\) being any \(+\) ve integer \()\) is (a) \(\frac{16}{n(2 n 1)} \mathrm{m} / \mathrm{s}\) (b) \(\frac{8}{\pi(2 n-1)} \mathrm{m} / \mathrm{s}\) (c) zero (d) \(\frac{16(2 n \mathrm{l})}{\pi} \mathrm{m} / \mathrm{s}\)
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