Problem 97 - Thomas Calculus

Question

Motion along a coordinate line A particle moves on a coordinate line with acceleration \(a=d^{2} s / d t^{2}=15 \sqrt{t}-(3 / \sqrt{t}),\) subject to the conditions that \(d s / d t=4\) and \(s=0\) when \(t=1 .\) Find a. the velocity \(v=d s / d t\) in terms of \(t\) b. the position \(s\) in terms of \(t\) .

Step-by-Step Solution

Answer

Velocity: \(v = 10t^{3/2} - 6t^{1/2}\); Position: \(s = 4t^{5/2} - 4t^{3/2}\).

Step 1: Integrate the Acceleration to Find Velocity

Given the acceleration function \(a = \frac{d^2s}{dt^2} = 15\sqrt{t} - \frac{3}{\sqrt{t}}\), we need to find the velocity \(v = \frac{ds}{dt}\). To do this, integrate the acceleration function with respect to time \(t\):\[ v = \int (15\sqrt{t} - \frac{3}{\sqrt{t}}) \, dt = \int 15t^{1/2} \, dt - \int 3t^{-1/2} \, dt. \]Evaluating these integrals:- \(\int 15t^{1/2} \, dt = 15 \cdot \frac{2}{3} t^{3/2} = 10t^{3/2}\),- \(\int -3t^{-1/2} \, dt = -3 \cdot 2t^{1/2} = -6t^{1/2}\).Thus, the velocity equation becomes:\[ v = 10t^{3/2} - 6t^{1/2} + C_1, \]where \(C_1\) is a constant of integration.

Step 2: Use Initial Condition to Find Constant for Velocity

We know that the velocity \(v = \frac{ds}{dt} = 4\) when \(t = 1\). Substitute these values into the velocity equation to solve for \(C_1\):\[ 4 = 10(1)^{3/2} - 6(1)^{1/2} + C_1. \]This simplifies to:\[ 4 = 10 - 6 + C_1, \]\[ 4 = 4 + C_1. \]Subtract 4 from both sides to find \(C_1\):\[ C_1 = 0. \]Thus, the velocity equation is \(v = 10t^{3/2} - 6t^{1/2}.\)

Step 3: Integrate the Velocity to Find Position

To find the position \(s\), integrate the velocity equation \(v = 10t^{3/2} - 6t^{1/2}\) with respect to \(t\):\[ s = \int (10t^{3/2} - 6t^{1/2}) \, dt = \int 10t^{3/2} \, dt - \int 6t^{1/2} \, dt. \]Evaluating these integrals:- \(\int 10t^{3/2} \, dt = 10 \cdot \frac{2}{5} t^{5/2} = 4t^{5/2}\),- \(\int -6t^{1/2} \, dt = -6 \cdot \frac{2}{3}t^{3/2} = -4t^{3/2}\).Thus, the position equation becomes:\[ s = 4t^{5/2} - 4t^{3/2} + C_2, \]where \(C_2\) is a constant of integration.

Step 4: Use Initial Condition to Find Constant for Position

We know that when \(t = 1\), \(s = 0\). Substitute these values into the position equation to solve for \(C_2\):\[ 0 = 4(1)^{5/2} - 4(1)^{3/2} + C_2. \]This simplifies to:\[ 0 = 4 - 4 + C_2, \]\[ 0 = 0 + C_2, \]So \(C_2 = 0. \)Thus, the position equation is \(s = 4t^{5/2} - 4t^{3/2}.\)

Key Concepts

Motion Along a Coordinate Line

Understanding motion along a coordinate line is crucial in physics and calculus. It describes how a particle or object moves back and forth along a straight path. Along a coordinate line, position, velocity, and acceleration are key concepts that help describe the motion of a particle. These are typically represented as functions of time.
In our exercise, we examine a particle with a given acceleration function, which changes as time increases. The goal is to determine how fast the particle is moving (its velocity) and where it is located along the coordinate line (its position), based on its acceleration and initial conditions.
Motion can be described in terms of:

• Position (\(s\)): Where the particle currently is on the coordinate line.
• Velocity (\(v\)): The speed and direction of the particle's movement.
• Acceleration (\(a\)): How the velocity changes over time.

The exercise gives us an acceleration function and requires us to find expressions for the velocity and position over time.

Integration

Integration is a powerful tool in calculus for finding accumulated values, such as area under a curve or total distance traveled. When dealing with motion, integration helps us move from acceleration to velocity, and then from velocity to position.
The exercise involves using integration to find the velocity of a particle given its acceleration. Integration essentially works by summing up the infinitesimal changes over an interval. In our case, the acceleration is given as a function of time, and by integrating this function, we obtain the velocity.
For example, if acceleration is \( a = 15\sqrt{t} - \frac{3}{\sqrt{t}} \), integrating with respect to time \( t \) gives the velocity function. This process involves calculating the integral:

• \( \int 15t^{1/2} \, dt = 10t^{3/2} \)
• \( \int -3t^{-1/2} \, dt = -6t^{1/2} \)

Adding these results will provide the velocity expression, including a constant of integration which is often determined using initial conditions.

Acceleration and Velocity

Acceleration describes how quickly the velocity of a particle changes. It can be constant, as in many basic problems, or can vary over time, as in our exercise. When acceleration isn't constant, it usually is given as a function of time, requiring calculus to understand its impacts on velocity and position.
Our task started by calculating velocity from acceleration, often done through integration. The relation between acceleration \( a \) and velocity \( v \) is given by \( v = \int a \, dt \). After integrating the acceleration equation, we get the velocity (e.g., \( v = 10t^{3/2} - 6t^{1/2} + C_1 \)) and need to use provided conditions to find the constant \( C_1 \).
Velocity is simply the rate of change of position. It shows not just how fast a particle is moving but in what direction. Once we have velocity, it can be further integrated to find position over time.

Initial Conditions

Initial conditions are vital in solving differential equations, particularly in motion problems. They help determine the specific solution to the general form provided by calculus. Without initial conditions, we have multiple possible solutions due to the constants that appear during integration.
In the exercise, an initial condition such as velocity at a specific time was given: \( \frac{ds}{dt} = 4 \) when \( t = 1 \). This information allows us to calculate the constant \( C_1 \) in the velocity equation. Similarly, when \( s = 0 \) at \( t = 1 \), it helps to find the constant \( C_2 \) in the position equation. These constants make our solution unique to the situation described.
To apply initial conditions:

• Substitute known values (e.g., time, velocity, position) into equations.
• Solve for the integration constants (e.g., \( C_1 \) and \( C_2 \)).

This ensures the mathematical model accurately reflects the real-world scenario outlined in the problem.