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Chapter 3: Problem 47

A particle moves along a straight line such that its position \(x\) at any time \(t\) is \(x=3 t^{2}-t^{3}\), where \(x\) is in metre and \(t\) in second, then: (a) at \(t=0\) acceleration is \(6 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(x\) -f curve has maximum at \(8 \mathrm{~m}\) (c) \(x\) -f curve has maximum at \(2 \mathrm{~s}\) (d) both (a) and (c) are correct

Short Answer

Expert verified
Option (d) both (a) and (c) are correct.

Step by step solution

01

Find the Velocity Function

The velocity function is the derivative of the position function with respect to time. Given the position function:\[ x = 3t^2 - t^3 \]Take the derivative:\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - t^3) = 6t - 3t^2 \]
02

Find the Acceleration Function

The acceleration function is the derivative of the velocity function with respect to time. Starting with the velocity function:\[ v(t) = 6t - 3t^2 \]Take the derivative:\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t - 3t^2) = 6 - 6t \]
03

Verify Acceleration at t = 0

Substitute \(t = 0\) into the acceleration function:\[ a(0) = 6 - 6(0) = 6 \]So, at \(t = 0\), the acceleration is \(6\, \text{m/s}^2\). This makes option (a) correct.
04

Identify the Maximum Point on Position-Time Curve

The maximum or minimum of the position function occurs where the velocity is zero. Set the velocity function to zero and solve for \(t\):\[ 6t - 3t^2 = 0 \]Factor out \(3t\):\[ 3t(2 - t) = 0 \]Thus, \(t = 0\) or \(t = 2\).
05

Determine Maximum Position Value

To confirm the maximum for the \(x\) function, calculate the position at \(t = 2\):\[ x(2) = 3(2)^2 - (2)^3 = 12 - 8 = 4 \]The position at \(t = 0\) is:\[ x(0) = 3(0)^2 - (0)^3 = 0 \]The maximum point for position is at \(t = 2\), with \(x = 4\, \text{m}\), not \(8\, \text{m}\). This makes option (b) incorrect and option (c) correct.

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Key Concepts

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

Position Function
In kinematics, understanding the position function is essential as it describes the location of a particle along a path at any given time. In our example, the position function is given by \[ x = 3t^2 - t^3 \] where \( x \) is the position in meters and \( t \) is the time in seconds. This quadratic-cubic equation tells us how the position of a particle changes over time.
  • The term \( 3t^2 \) represents a parabolic motion indicating upward movement.
  • The \( -t^3 \) term suggests a slowing down effect as time progresses, due to the negative cubic component.
To find specific positions at given times, substitute the time value into the function. For instance, at \( t=2 \) seconds, plugging the value into the function will give the position of the particle at that specific time.
Velocity Function
The velocity function provides insights into how the speed of the particle changes over time. It's obtained by differentiating the position function with respect to time. For our problem:\[ v(t) = \frac{dx}{dt} = 6t - 3t^2 \]This represents a linear-quadratic relationship:
  • The linear term \( 6t \) signifies a constant initial acceleration.
  • The quadratic term \( -3t^2 \) indicates that as time increases, the velocity starts to decrease due to deceleration from the cubic term.
To find when the particle stops (when its velocity is zero), solve for \( t \) by setting:\[ 6t - 3t^2 = 0 \]This gives critical time values where velocity might be zero, helping us find potential maximum or minimum points on the position-time graph.
Acceleration Function
Acceleration describes the rate of change of velocity over time. It's found by differentiating the velocity function. In our case, the acceleration function is:\[ a(t) = \frac{dv}{dt} = 6 - 6t \]This linear expression simplifies understanding of acceleration dynamics:
  • The constant term \( 6 \) indicates the initial acceleration at \( t = 0 \).
  • The \( -6t \) term shows that as time increases, the overall acceleration decreases linearly.
To verify acceleration at a specific time, substitute the time value into the function. For example, at \( t=0 \), the acceleration is \( 6 \text{ m/s}^2 \), consistent with the initial given condition.
Maximum and Minimum Points
Maximum and minimum points on the position-time curve help identify crucial moments in the particle's journey, like peak position or turning points. These extreme points occur when the velocity is zero because at these points, the rate of position change stops temporarily.
To determine these points, solve the equation:\[ 6t - 3t^2 = 0 \]Factoring it gives:\[ 3t(2 - t) = 0 \]yielding two solutions \( t=0 \) and \( t=2 \).
  • Plugging \( t=2 \) into the position function provides \( x = 4 \text{ m} \), confirming a local maximum in this scenario.
  • The solution illustrates that the maximum height reached is \( x=4 \text{ m} \) at \( t=2 \text{ seconds} \), thereby proving the option about a peak position's timing as correct.
Recognizing these points is vital for analyzing motion in kinematics, as it helps visualize and plan around peak and transition moments in movement.

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Most popular questions from this chapter

A ball is dropped vertically from a height \(d\) above the ground. It hits the ground and bounces up vertically to a height \(d / 2 .\) Neglecting subsequent motion and air resistance, its speed \(v\) varies with the height \(h\) above the ground as :

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