Problem 105 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Given Information

We are given the acceleration function of a particle as \( a(t) = \frac{d^2 s}{d t^2} = 15\sqrt{t} - \frac{3}{\sqrt{t}} \). We also have initial conditions: \( \frac{d s}{d t} = 4 \) and \( s = 0 \) when \( t = 1 \). The task is to find expressions for velocity \( v(t) \) and position \( s(t) \).

Step 2: Integrate Acceleration to Find Velocity

To find the velocity \( v(t) = \frac{d s}{d t} \), integrate the acceleration \( a(t) \) with respect to \( t \):\[ v(t) = \int a(t) \, dt = \int \left(15\sqrt{t} - \frac{3}{\sqrt{t}}\right) dt \]Breaking it into parts:\[ v(t) = \int 15t^{1/2} \, dt - \int 3t^{-1/2} \, dt \]Let's solve these integrals individually:\[ \int 15t^{1/2} \, dt = 15 \cdot \frac{2}{3} t^{3/2} = 10t^{3/2} \]\[ \int 3t^{-1/2} \, dt = 3 \cdot 2t^{1/2} = 6t^{1/2} \]Thus, the primitive function is:\[ v(t) = 10t^{3/2} - 6t^{1/2} + C \] where \( C \) is the constant of integration.

Step 3: Apply Initial Condition for Velocity

Use the initial condition \( v(1) = 4 \) to find the constant \( C \):\( 4 = 10(1)^{3/2} - 6(1)^{1/2} + C \)That simplifies to:\[ 4 = 10 - 6 + C \]\[ C = 4 - 4 = 0 \]Thus, \( v(t) = 10t^{3/2} - 6t^{1/2} \).

Step 4: Integrate Velocity to Find Position

Next, integrate the velocity \( v(t) = 10t^{3/2} - 6t^{1/2} \) to get the position function \( s(t) \):\[ s(t) = \int (10t^{3/2} - 6t^{1/2}) \, dt \]Solve these integrals:\[ \int 10t^{3/2} \, dt = 10 \cdot \frac{2}{5}t^{5/2} = 4t^{5/2} \]\[ \int 6t^{1/2} \, dt = 6 \cdot \frac{2}{3} t^{3/2} = 4t^{3/2} \]So:\[ s(t) = 4t^{5/2} - 4t^{3/2} + D \] where \( D \) is the constant of integration.

Step 5: Apply Initial Condition for Position

Use the initial condition \( s(1) = 0 \) to find \( D \):\( 0 = 4(1)^{5/2} - 4(1)^{3/2} + D \)That simplifies to:\[ 0 = 4 - 4 + D \]\[ D = 0 \]Thus, \( s(t) = 4t^{5/2} - 4t^{3/2} \).

Key Concepts

Acceleration Function

Acceleration is a measure of how the velocity of an object changes over time. In our problem, we are given a specific acceleration function for a particle moving along a coordinate line, represented by the equation: \[ a(t) = 15\sqrt{t} - \frac{3}{\sqrt{t}} \]This function shows how acceleration depends on time \( t \). Let's break it down:

• The term \( 15\sqrt{t} \) signifies that acceleration increases with the square root of time, suggesting rapid acceleration initially that slows down as time progresses.
• The term \( -\frac{3}{\sqrt{t}} \) indicates a diminishing component as time increases, meaning this factor reduces the overall acceleration especially when \( t \) is small.

This setup points out the dynamic nature of particle movement on a coordinate line, beginning with a higher start and leveling out as time passes.

Velocity Integration

To determine the velocity function \( v(t) \), we integrate the given acceleration function over time. Integrating is the process of finding a function whose derivative equals the original function, thereby providing us with velocity:\[ v(t) = \int a(t) \, dt = \int \left(15\sqrt{t} - \frac{3}{\sqrt{t}}\right) dt \]Breaking integration into parts, we have two fundamental integrals:

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

Combine these to get the velocity equation:\[ v(t) = 10t^{3/2} - 6t^{1/2} + C \]Integration yields a function that describes velocity over time, with \( C \) representing a constant determined by initial conditions.

Initial Conditions

When dealing with differential equations, initial conditions are crucial to finding specific solutions. Given initial values help us find particular constants that arise during integration. In our exercise, we had:

• The initial velocity, \( v(1) = 4 \), which means that at time \( t = 1 \), the velocity is 4 units.
• The initial position, \( s(1) = 0 \), indicating that at \( t = 1 \), the position along the coordinate line is 0 units.

These conditions allow us to solve for our constants during integration:1. Using \( v(1) = 4 \), we solved for the constant \( C \) in the velocity function. Substituting in the initial conditions, \( C \) turned out to be 0.2. Similarly, applying \( s(1) = 0 \) in the position function determined that the constant \( D \) is also 0.These steps ensure that our integrated functions precisely model the system described by the problem.

Position Function

The position function \( s(t) \) tells us where our particle is located along the coordinate line at any time \( t \). Obtaining this involves integrating the velocity function:\[ s(t) = \int (10t^{3/2} - 6t^{1/2}) \, dt \]We solve the integrals as follows:

• \( \int 10t^{3/2} \, dt = 4t^{5/2} \)
• \( \int 6t^{1/2} \, dt = 4t^{3/2} \)

Combine to get:\[ s(t) = 4t^{5/2} - 4t^{3/2} + D \]Using the initial condition \( s(1) = 0 \) helped us determine that \( D = 0 \). Thus, the final position function becomes:\[ s(t) = 4t^{5/2} - 4t^{3/2} \]This expression shows how the particle's position changes over time based on the given acceleration.