Problem 32 - Thomas Calculus

Question

Exercises \(31-34\) give the position function \(s=f(t)\) of a body moving along the \(s\) -axis as a function of time \(t .\) Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t)\) . Comment on the body's behavior in relation to the signs and values of \(v\) and \(a\) . Include in your commentary such topics as the following: a. When is the body momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$ s=t^{2}-3 t+2, \quad 0 \leq t \leq 5 $$

Step-by-Step Solution

Answer

The body is at rest at \(t=\frac{3}{2}\); moves right for \(t>\frac{3}{2}\) and left otherwise; changes direction at \(t=\frac{3}{2}\); speeds up for \(t>\frac{3}{2}\) and slows down for \(t<\frac{3}{2}\); moves slowest at \(t=\frac{3}{2}\) and farthest from the origin at \(t=5\).

Step 1: Find the Velocity Function

The velocity function is the first derivative of the position function. For \(s(t) = t^2 - 3t + 2\), we find the derivative \(v(t) = \frac{ds}{dt} = 2t - 3\).

Step 2: Find the Acceleration Function

The acceleration function is the derivative of the velocity function. So, \(a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} = 2\).

Step 3: Determine When the Body is at Rest

The body is momentarily at rest when the velocity is zero. Set \(v(t) = 0\): \[2t - 3 = 0\] Solve for \(t\): \[t = \frac{3}{2}\]

Step 4: Determine Direction of Movement

The body moves to the right (up) when \(v(t) > 0\) and to the left (down) when \(v(t) < 0\). \(v(t) = 2t - 3\): - Right when \(t > \frac{3}{2}\) - Left when \(t < \frac{3}{2}\)

Step 5: Identify Where the Body Changes Direction

A change in direction occurs when the velocity changes sign, which happens when \(v(t) = 0\) at \(t = \frac{3}{2}\).

Step 6: Analyze Speeding Up and Slowing Down

The body speeds up when velocity and acceleration have the same sign. - For \(t > \frac{3}{2}\), both \(v > 0\) and \(a = 2 > 0\). - Slows down when velocity and acceleration have opposite signs, which occurs for \(t < \frac{3}{2}\), when \(v < 0\) and \(a = 2 > 0\).

Step 7: Find When the Body is Moving Fastest and Slowest

The body moves fastest and slowest based on the speed, \(|v(t)|\). Compute this before and after \(t = \frac{3}{2}\) to determine speed. Given constant acceleration, theoretically, body moves slowest at \(t = \frac{3}{2}\) due to direction change.

Step 8: Determine Maximum Displacement from Origin

Evaluate \(s(t)\) at critical points and boundaries (\(t = 0, \frac{3}{2}, 5\)) to determine farthest point from the origin: \(s(0) = 2\), \(s\left(\frac{3}{2}\right) = -\frac{1}{4}\), \(s(5) = 12\). Closest to origin at \(t = \frac{3}{2}\); farthest at \(t = 5\).

Key Concepts

Position Function

In kinematics, the position function describes where a moving object is located along a path at any given time. It defines the relationship between time and the position of the object in a linear space. For example, the position function given as \[ s(t) = t^2 - 3t + 2 \]provides a mathematical way to predict the object's position at any moment from time \( t = 0 \) to \( t = 5 \). This function is essential for understanding how the position of the object changes over time.
To graph this function, evaluate it at different \( t \) values. This visualization helps in observing key characteristics like maximum and minimum positions, which are crucial for deeper analysis of motion behaviors. By examining how the function changes, one can also infer patterns in movement such as sinusoidal or linear trends.

Velocity Function

The velocity function gives insight into how fast an object is moving and in which direction. It is derived from the position function by calculating the first derivative. For our example, we derive the velocity as \[ v(t) = rac{ds}{dt} = 2t - 3. \]
This tells us the rate of change of position over time, allowing to predict when and where the object moves faster or slower.
Key points to note:

• The object is momentarily at rest when \( v(t) = 0 \), which happens at \( t = \frac{3}{2} \).
• The object moves right or up (forward) when \( v(t) > 0 \) and left or down (backward) when \( v(t) < 0 \).
• A change in sign of \( v(t) \) indicates a change in direction.

Understanding velocity helps determine critical moments in motion, such as when the object ceases movement or picks up speed again.

Acceleration Function

Acceleration is the measure of how quickly the velocity changes, derived from taking the second derivative of the position function, or the first derivative of the velocity function. For this exercise, the acceleration is constant:\[ a(t) = rac{dv}{dt} = 2. \]
A positive acceleration consistently pushes the object further in the same direction of its initial movement once it achieves positive velocity.
This consistent acceleration means:

• The object speeds up when both its velocity and acceleration share the same sign. For \( t > \frac{3}{2} \), the velocity and acceleration are both positive, leading to speeding up.
• It slows down when the signs are opposite, observed for \( t < \frac{3}{2} \), where the object has negative velocity but positive acceleration.

Analyzing acceleration provides insight into how motion dynamics will unfold over time, showing force interactions and changes in movement potential.

Motion Analysis

Motion analysis combines the insights from position, velocity, and acceleration functions to paint a complete picture of an object's behavior. By examining these functions together, one can fully understand the object's path and pivotal motion events.
Here's a quick rundown:

• The body starts at \( t = 0 \) and is farthest from the origin at \( t = 5 \), with the position value highest at 12.
• The object changes direction at \( t = \frac{3}{2} \) due to the velocity crossing zero.
• It speeds up post \( t = \frac{3}{2} \), moving fastest as t approaches 5.
• Though slowest at the moment it changes direction due to a temporary zero velocity, it regains speed thanks to constant acceleration.

This comprehensive analysis helps predict future motion, plan interventions, or adjust parameters for desired paths in various applications. Understanding motion also reveals the impact of initial conditions set in the position function and how they translate through velocity and acceleration.