Problem 14 - Calculus for Biology and Medicine

Question

14\. Velocity Suppose a particle moves along a straight line. The position at time \(t\) is given by $$ s(t)=3 t-t^{2}, \quad t \geq 0 $$ where \(t\) is measured in seconds and \(s(t)\) is measured in meters. (a) Graph \(s(t)\) for \(t \geq 0\). (b) Use the graph in (a) to answer the following questions: (i) Where is the particle at time \(0 ?\) (ii) Is there another time at which the particle visits the location where it was at time \(0 ?\) (iii) How far to the right on the straight line does the particle travel? (iv) How far to the left on the straight line does the particle travel? (v) Where is the velocity positive? negative? equal to \(0 ?\) (c) Find the velocity of the particle. (d) When is the velocity of the particle equal to \(1 \mathrm{~m} / \mathrm{s}\) ?

Step-by-Step Solution

Answer

The particle starts at 0 meters, returns there at 3 seconds, travels up to \( \frac{9}{4} \) meters right, never further left than 0 meters, with velocity zero at 1.5 seconds. Velocity is 1 m/s at 1 second.

Step 1: Graph the Position Function

To graph the position function \( s(t) = 3t - t^2 \), find several values of \( s(t) \) for different \( t \) values starting from \( t=0 \). You might choose \( t=0, 1, 2, 3 \), and so forth, calculate their corresponding \( s(t) \), and plot these points on a graph. The curve should be a downward-opening parabola as it represents a quadratic function with a negative coefficient for \( t^2 \).

Step 2: Initial Position of the Particle

Evaluate \( s(t) \) at \( t = 0 \): \( s(0) = 3(0) - (0)^2 = 0 \). Thus, the particle is at position 0 meters at time zero.

Step 3: Return to Initial Position

To find if there is another \( t \) for which \( s(t) = 0 \), solve: \[ 3t - t^2 = 0 \] Factoring gives \( t(3 - t) = 0 \), so \( t = 0 \) or \( t = 3 \). Thus, the particle returns to the initial position at \( t = 3 \).

Step 4: Maximum Rightward Position

To find how far to the right the particle travels, calculate the vertex of the parabola, which is the maximum point here. The vertex \( t \) is given by \( t = -\frac{b}{2a} \) where \( a = -1, b = 3 \), resulting in \( t = \frac{3}{2} \). Calculate \( s(t) \) at \( t = \frac{3}{2} \): \[ s \left( \frac{3}{2} \right) = 3 \left( \frac{3}{2} \right) - \left( \frac{3}{2} \right)^2 = \frac{9}{2} - \frac{9}{4} = \frac{9}{4} \] Thus, the maximum rightward journey is \( \frac{9}{4} \) meters.

Step 5: Leftward Limits

The particle returns to the starting position \( s = 0 \) from its maximum right position, hence it does not travel left of \( 0 \) meters.

Step 6: Determine Velocity

The velocity \( v(t) \) is the derivative of \( s(t) \). Differentiate: \[ v(t) = \frac{d}{dt} [3t - t^2] = 3 - 2t \] This formula gives the velocity at any time \( t \).

Step 7: Identify Velocity Conditions

- Velocity is positive where \( 3 - 2t > 0 \): \( t < \frac{3}{2} \).- Velocity is negative where \( 3 - 2t < 0 \): \( t > \frac{3}{2} \).- Velocity is zero where \( 3 - 2t = 0 \): \( t = \frac{3}{2} \).

Step 8: When Velocity is 1 m/s

Set the velocity equation to \( 1 \) and solve for \( t \): \[ 3 - 2t = 1 \] Solving gives \( 2t = 2 \), so \( t = 1 \). Thus, the velocity is \( 1 \text{ m/s} \) at \( t = 1 \).

Key Concepts

velocity

Velocity in calculus refers to the rate of change of a particle's position with respect to time. It tells us how fast and in which direction the particle is moving.
This is vital for understanding a particle's movement over time.Let's break it down:

• Velocity as a Derivative: The velocity of a particle is the derivative of its position function with respect to time. It mathematically represents the slope of the position curve.
• Positive and Negative Velocity: Positive velocity indicates movement in one direction (commonly the right or upwards), while negative velocity suggests movement in the opposite direction.
• Instantaneous Velocity: This is the velocity of a particle at any specific point in time.

In our example, the velocity function derived from the position function is given by \[ v(t) = 3 - 2t \] Here, you can find instantaneous velocities for different values of time \( t \). Understanding velocity helps in predicting the future position and speed of the particle.

position function

A position function, in simple terms, is a mathematical equation that tells us the location of a particle at any given time. This is crucial for tracking where the particle is over time.With the formula, \[ s(t) = 3t - t^2 \] it gives the position at any time \( t \), where \( s(t) \) is measured in meters and \( t \) in seconds.What does this mean practically?

• The position at time zero \( t=0 \) helps us know the starting point, which is crucial for mapping the path.
• By solving for different times as in the example (finding \( s(t)=0 \)), we can discover when the particle returns to its initial position.
• The shape of the graph (a parabola here) visually represents the directional travel of the particle.

This function points out where the particle is on the line at any given moment, critical for analyzing its movement.

derivative

The derivative is the backbone of calculus and is used to reveal many physical concepts, including motion. It's about finding the rate at which one quantity changes in relation to another. In our context, it's the rate of change of position, or simply the velocity.Using the position function \[ s(t) = 3t - t^2 \] its derivative, the velocity function, is determined as \[ v(t) = \frac{d}{dt} (3t - t^2) = 3 - 2t \].Why is it so important?

• The process of differentiation allows us to find instantaneous changes — how fast something is changing at an exact point. This especially matters in predicting and understanding motion.
• From the derivative, we not only deduce velocity but can identify acceleration (derivative of velocity) or further changes in motion.
• The derivative's role in understanding the maxima and minima points on a curve aids identifying critical points like maximum displacement in particle motion.

Derivatives hugely impact how we interpret functions by dissecting their behaviors at specific points in their domain.

particle motion

Particle motion outlines the path taken by a particle as it moves along a specific course. This includes not only its position but also how quickly and in what direction it's changing at any given moment.This concept unites both position and velocity functions. When we explore particle motion with the given position function \[ s(t) = 3t - t^2 \]it reveals a few insights into movement:

• Direction of Movement: By analyzing velocity sign changes (such as where the derivative is zero), it tells us when the particle switches direction.
• Travel Path: The path includes areas of rest (where velocity is zero) and checks areas visited once or repetitively.
• Distance Travelled: Investigating maximums and minimums can uncover how far to the right or left the particle has moved from an initial position.

The study of particle motion further involves graphing the position function, determining areas of acceleration or deceleration, and deducing distance and displacement.Understanding particle motion gives a comprehensive view of an object's journey, invaluable in real-world applications from vehicle tracking to physics simulations.