Problem 20 - Fundamentals of Physics

Question

(a) If the position of a particle is given by \(x=20 t-5 t^{3}\), where \(x\) is in meters and \(t\) is in seconds, when, if ever, is the particle's velocity zero? (b) When is its acceleration \(a\) zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph \(x(t), v(t),\) and \(a(t)\)

Step-by-Step Solution

Answer

(a) Velocity is zero at \(t = \pm\sqrt{\frac{4}{3}}\). (b) Acceleration is zero at \(t=0\). (c) \(a<0\) for \(t>0\). (d) \(a>0\) for \(t<0\). (e) Graphs depict these functions' behaviors over time.

Step 1: Identify and differentiate position function to find velocity

The position of the particle is given as a function of time: \(x(t) = 20t - 5t^3\). To find the velocity \(v(t)\), differentiate this function with respect to time \(t\):\[v(t) = \frac{dx}{dt} = \frac{d}{dt} (20t - 5t^3) = 20 - 15t^2.\]

Step 2: Solve for zero velocity

Set the velocity function \(v(t)\), obtained in Step 1, to zero to find when the velocity is zero:\[20 - 15t^2 = 0.\]Solve for \(t\):\[15t^2 = 20 \quad \Rightarrow \quad t^2 = \frac{20}{15} = \frac{4}{3}.\]Thus, \(t = \sqrt{\frac{4}{3}}\) or \(t = -\sqrt{\frac{4}{3}}\).

Step 3: Differentiate velocity function to find acceleration

Differentiate the velocity function \(v(t) = 20 - 15t^2\) with respect to time to obtain the acceleration \(a(t)\):\[a(t) = \frac{dv}{dt} = -30t.\]

Step 4: Determine when acceleration is zero

Set the acceleration function \(a(t) = -30t\) to zero and solve for \(t\):\[-30t = 0 \quad \Rightarrow \quad t = 0.\]

Step 5: Identify time range for negative acceleration

Acceleration \(a(t) = -30t\) is negative when \(-30t < 0\), leading to \(t > 0\). Thus, for positive \(t\), acceleration is negative.

Step 6: Identify time range for positive acceleration

The acceleration \(a(t) = -30t\) becomes positive when \(-30t > 0\), which results in \(t < 0\). Thus, for negative \(t\), acceleration is positive.

Step 7: Graph functions for x(t), v(t), a(t)

- Graph \(x(t) = 20t - 5t^3\) shows a cubic function that changes direction.- Graph \(v(t) = 20 - 15t^2\) shows a quadratic curve opening downwards, crossing the t-axis where velocity is zero.- Graph \(a(t) = -30t\) is a linear function declining through the origin, indicating how acceleration changes from positive to negative.

Key Concepts

particle motion

Particle motion refers to the study of how particles move through space over time. In this context, a particle is an object considered as a single point with respect to motion analysis. Understanding particle motion involves analyzing various physical quantities such as position, velocity, and acceleration. These quantities reveal how an object's location evolves, how fast it moves, and how its speed changes.

In kinematics, the focus is on understanding this motion without considering the forces behind it. By studying particle motion, we can predict where an object will be at a certain time, determine points where its velocity is zero, and assess intervals during which it speeds up or slows down. This deep understanding builds the foundation for more complex motion studies and practical applications.

position function

The position function, often denoted as \(x(t)\), describes the location of a particle as a function of time. It maps the time \(t\) to a spatial position \(x\). For example, the position function \(x = 20t - 5t^3\) indicates how a particle's position progresses over time.

In this specific function, each term represents aspects of the particle's motion:

• The term \(20t\) suggests a straight-line motion, assuming no other factors changing over time.
• The \(-5t^3\) term reflects changes in motion direction and acceleration, indicating the path is not linear.

By analyzing this position function, we can gain insights, such as calculating specific moments in time when the velocity is zero, given its derivative forms.

velocity and acceleration

Velocity and acceleration are crucial components when examining how a particle's motion evolves. Velocity represents the rate at which an object's position changes over time and is obtained by differentiating the position function:

\[v(t) = \frac{dx}{dt} = 20 - 15t^2\]

This velocity function shows how fast the particle is moving and in which direction. Roots of this function, such as \(t = \pm \sqrt{\frac{4}{3}}\), indicate moments when the particle momentarily stops.

Acceleration, on the other hand, is the rate of change of velocity. It is the derivative of the velocity function:

\[a(t) = \frac{dv}{dt} = -30t\]

This expression explains how quickly the particle's speed changes. Notably, acceleration is zero when \(t = 0\), meaning any change in velocity is momentarily paused. Additionally, analyzing intervals where acceleration is positive or negative can reveal phases of speeding up and slowing down.

graphing motion functions

Graphing motion functions like position, velocity, and acceleration provide visual insights into particle motion. These graphs illustrate the relationship between time and each respective kinematic property.

• The position graph \(x(t) = 20t - 5t^3\) displays a cubic curve, indicating points where direction changes, characterized by inflection and turning points.


• The velocity graph \(v(t) = 20 - 15t^2\) resembles a downward-opening parabola, illustrating points where the velocity goes to zero, marking changes in direction.


• The acceleration graph \(a(t) = -30t\) is a straight line declining through the origin, showing a continuous decrease in acceleration, crossing the time-axis at zero.

Visual representations of these functions allow students to better understand and predict the nature of motion over time, reinforcing theoretical calculations.