Problem 36 - University Calculus: Early Transcendentals

Question

Give the position function \(s=f(t)\) of an object 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 object'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 object 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? $$s=4-7 t+6 t^{2}-t^{3}, \quad 0 \leq t \leq 4$$

Step-by-Step Solution

Answer

The object is at rest at \(t=1\) and \(t=2.33\). It moves right in \((0,1)\) and \((2.33,4)\). It changes direction at \(t=1, 2.33\). Maximum speed is at \(t=0\), farthest at \(t=4\).

Step 1: Find the Velocity Function

The velocity function is the derivative of the position function. Taking the derivative of \(s(t) = 4 - 7t + 6t^2 - t^3\), we get:\[v(t) = \frac{ds}{dt} = -7 + 12t - 3t^2\]

Step 2: Find the Acceleration Function

The acceleration function is the derivative of the velocity function. Taking the derivative of \(v(t) = -7 + 12t - 3t^2\), we get:\[a(t) = \frac{d^2s}{dt^2} = 12 - 6t\]

Step 3: Analyze Critical Points for Velocity

To find when the object is momentarily at rest, set \(v(t) = 0\):\[-7 + 12t - 3t^2 = 0\]This simplifies to:\[3t^2 - 12t + 7 = 0\]Using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find the critical points. Here, \(a=3\), \(b=-12\), \(c=7\). Solving yields:\[t = \frac{4 \pm \sqrt{4^2 - 3 \cdot 7}}{2} = 1, 2.33\]

Step 4: Determine Direction of Movement

The object moves to the right when \(v(t) > 0\) and to the left when \(v(t) < 0\). Evaluate the sign of \(v(t)\) around the critical points to find the intervals.- For \(0 < t < 1\), \(v(t) > 0\) (moving right).- For \(1 < t < 2.33\), \(v(t) < 0\) (moving left).- For \(2.33 < t < 4\), \(v(t) > 0\) (moving right again).

Step 5: Determine Changes in Direction and Speed

The object changes direction at moments where \(v(t) = 0\) turns from positive to negative or vice-versa: \(t = 1\) and \(t = 2.33\).- Speed changes depend on the sign of \(a(t)\): - Speeding up when \(v(t)\) and \(a(t)\) have the same sign. - Slowing down when they have opposite signs.Calculate \(a(t)\) at critical points and intervals to analyze speed:- \(a(t) > 0\) for \(t < 2\), \(v(t) < 0\) implies slowing down in \((1, 2)\) and speeding up in \((2, 2.33)\).

Step 6: Determine Maximum and Minimum Speeds

Find extreme values of speed by evaluating the magnitude of velocity at critical points and boundary:- Maximum speed is at \(t = 0\), because velocity is max positive.- Minimum speed is at \(t = 2.33\), because velocity is min negative.

Step 7: Determine Farthest Distance from Origin

Evaluate the position function \(s(t)\) at critical points and verify:- Find \(s(0), s(1), s(2.33), s(4)\) to determine the largest absolute value:- At each point, \(s(0) = 4, s(1)=2, s(2.33)=4.59, s(4)=-8\), so the farthest point • considering absolute distance from the origin is \(s(4) = -8\).

Key Concepts

Position Function

The position function describes how the location of an object changes over time. In our exercise, the position function is given as: \[ s(t) = 4 - 7t + 6t^2 - t^3 \] This formula indicates the position \(s\) of the object on the \(s\)-axis as it moves over time \(t\). Each term in this polynomial equation contributes to the overall trajectory of the object.

• The constant term, \(4\), represents the initial position of the object when \(t = 0\).
• The linear term, \(-7t\), adds an initial backward movement.
• The quadratic term, \(6t^2\), affects the object's movement by causing it to speed up in the positive direction as time increases.
• The cubic term, \(-t^3\), eventually overpowers other terms, reversing the motion as \(t\) grows large due to its negative coefficient, showcasing a change of direction.

Understanding how these components interact helps visualize the path taken by the object as it moves along the \(s\)-axis.

Velocity Function

The velocity function is critical for understanding how fast and in what direction the object is moving at any given moment. It's the first derivative of the position function. For our object:\[ v(t) = \frac{ds}{dt} = -7 + 12t - 3t^2 \] This derivative informs us of the object's speed and direction at various times. Evaluating \(v(t)\) helps determine:

• When the object is stationary: This occurs when \(v(t) = 0\), yielding critical points at \(t = 1\) and \(t = 2.33\).
• The direction of movement: Positive \(v(t)\) values indicate movement to the right; negative values signify movement to the left.

Such analysis provides insight into when the object changes direction, which happens at points where velocity crosses the zero mark in its path.

Acceleration Function

Acceleration measures how the velocity changes over time, essential for understanding the object's change in motion. It is the derivative of the velocity function. In our example:\[ a(t) = \frac{d^2s}{dt^2} = 12 - 6t \]Here, the acceleration tells us:

• Speeding up: When the signs of \(v(t)\) and \(a(t)\) match, the object is accelerating in the direction of motion.
• Slowing down: Opposite signs of \(v(t)\) and \(a(t)\) indicate deceleration.

For \(t < 2\), \(a(t) > 0\); thus, acceleration works in the direction the object is moving. Post \(t = 2\), \(a(t) < 0\), suggesting a shift in this relationship, affecting whether the object speeds up or slows down. This critical information helps in understanding intervals and moments of acceleration or deceleration.

Critical Points Analysis

Critical points occur where the derivative, in this case, the velocity function \(v(t)\), equals zero, indicating a potential change in direction or speed. Solving \(v(t) = 0\) in our exercise yields critical points \(t = 1\) and \(t = 2.33\). Each critical point:

• Marks a possible change in direction.
• Helps determine intervals where the object either moves to the left or right.
• Informs when the object is momentarily at rest.
• Frames context for analyzing speed changes, as crossing these points implies a switch in movement behavior dictated by acceleration.

Evaluating the position, velocity, and acceleration functions around these critical points helps piece together the narrative of the object's journey along the \(s\)-axis. This analysis is pivotal for grounding the study of motion calculus, essential for both theoretical and practical applications in physics and engineering.