Problem 101 - Thomas Calculus in SI Units

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

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

Step 1: Find the Velocity Function

To find the velocity function, integrate the acceleration function with respect to time. Start with the given acceleration function: \( a = \frac{d^2 s}{d t^2} = 15 \sqrt{t} - \frac{3}{\sqrt{t}} \). Integrate to find velocity \( v = \int a \, dt \). Separate the terms and integrate:\[ v(t) = \int \left( 15 \sqrt{t} \right) \, dt - \int \left( \frac{3}{\sqrt{t}} \right) \, dt \]Calculate each integral separately. For the first integral:\[ \int 15 \sqrt{t} \, dt = 15 \cdot \frac{2}{3} t^{3/2} = 10 t^{3/2} \]For the second integral:\[ \int \frac{3}{\sqrt{t}} \, dt = 3 \cdot 2 t^{1/2} = 6 t^{1/2} \]Thus, the velocity is:\[ v(t) = 10 t^{3/2} - 6 t^{1/2} + C \]To find the constant \( C \), use the condition \( v(1) = 4 \). Substitute \( t = 1 \):\[ 4 = 10(1)^{3/2} - 6(1)^{1/2} + C \]\[ 4 = 10 - 6 + C \]\[ C = 0 \]So, the velocity function is \( v(t) = 10 t^{3/2} - 6 t^{1/2} \).

Step 2: Find the Position Function

To find the position \( s \) in terms of \( t \), integrate the velocity function. The velocity function is \( v(t) = \frac{d s}{d t} = 10 t^{3/2} - 6 t^{1/2} \). Integrate to find position \( s = \int v(t) \, dt \):\[ s(t) = \int \left( 10 t^{3/2} - 6 t^{1/2} \right) \, dt \]This gives:\[ s(t) = 10 \cdot \frac{2}{5} t^{5/2} - 6 \cdot \frac{2}{3} t^{3/2} \]\[ s(t) = 4 t^{5/2} - 4 t^{3/2} + C \]Use the initial condition \( s(1) = 0 \) to find \( C \):\[ 0 = 4(1)^{5/2} - 4(1)^{3/2} + C \]\[ 0 = 4 - 4 + C \]\[ C = 0 \]Hence, the position function is \( s(t) = 4 t^{5/2} - 4 t^{3/2} \).

Key Concepts

Integration in Calculus

Integration is a fundamental concept in calculus, often used to find quantities such as areas, volumes, and, in this context, the antiderivatives of a function. When dealing with motion, integration allows us to move from an object's acceleration function to its velocity function, and further to its position function.

In the given exercise, the acceleration function is defined as:

• \[ a = 15 \sqrt{t} - \frac{3}{\sqrt{t}} \]

To determine the velocity, we need to integrate the acceleration function with respect to time, \(t\). This means we find a function whose derivative gives the acceleration.

The concept of finding an antiderivative is crucial here. For example, integrating \( \sqrt{t} \) results in \( \frac{2}{3}t^{3/2} \), which is illustrated in the solution as part of the integral:

• \[ \int 15 \sqrt{t} \, dt = 15 \cdot \frac{2}{3} t^{3/2} = 10 t^{3/2} \]

The integration process shows how reversing differentiation retrieves the original function (plus a constant of integration, \(C\)).

This process is essential to finding velocity and position when only acceleration is known. Through integration, we're able to trace back through the effects of time and initial conditions to get full expressions for velocity and position.

Motion Along a Line

Understanding motion along a line requires considering kinematic concepts such as position, velocity, and acceleration. Each describes different aspects of a particle’s movement along a one-dimensional path.

In many calculus problems, objects are moving in a straight line, and you need to determine velocities or positions at various times. This is much like in our exercise, where a particle is moving along a coordinate line.

Let's break down the motion:

• Position: indicates where the particle is on the line; defined mathematically as \(s(t)\).
• Velocity: is the rate of change of position concerning time; hence it's \(\frac{ds}{dt}\).
• Acceleration: is the rate at which the velocity changes; represented as \(\frac{d^2s}{dt^2}\).

Motion in a straight path simplifies calculations as it focuses solely on how these quantities change over time due to linear constraints.

By evaluating the integral of the particle's known acceleration, you determine its change in speed over time. This lets you visualize motion in a temporal sequence, highlighting how constants like initial conditions shape trajectories.

Acceleration and Velocity

Acceleration and velocity are two closely related but distinct concepts in motion dynamics. They are often core concepts examined using calculus.

Acceleration describes how velocity changes over time. It can be expressed in terms of differentials as \( a = \frac{d^2 s}{d t^2} \), which indicates it's the derivative of the velocity. The given problem has acceleration:

• \[ a(t) = 15 \sqrt{t} - \frac{3}{\sqrt{t}} \]

Through the process of integration, acceleration leads to the velocity expression, \( v(t) = 10 t^{3/2} - 6 t^{1/2} \), by retracing our steps from change in speed to speed itself.

Velocity, on the other hand, portrays the rate of change of the position over time. It's calculated via the integral of the acceleration function, making it contingent upon the starting conditions specified. Here, knowing \( v(1)=4 \) allows us to solve for the integration constant, signaling the initial velocity at a specified moment.

Overall, the differences in these concepts emphasize their interdependence in characterizing motion. Calculus connects them in a manner that allows generation and solving of function models usable in physics and engineering applications.