Problem 24 - The Calculus with Analytic Geometry

Question

In Exercises 24 through 27 , a particle is moving along a straight line according to the given equation of motion, where \(s\) \(\mathrm{ft}\) is the directed distance of the particle from the origin at \(t \mathrm{sec}\). Find the time when the instantaneous acceleration is zero, and then find the directed distance of the particle from the origin and the instantaneous velocity at this instant. $$ s=2 t^{3}-6 t^{2}+3 t-4, t \geq 0 $$

Step-by-Step Solution

Answer

The acceleration is zero at \( t = 1 \) sec. The distance from the origin is -5 feet and the instantaneous velocity is -3 ft/sec at this time.

Step 1: Find the velocity

The velocity is the first derivative of the position function. Given the position function, \( s(t) = 2t^3 - 6t^2 + 3t - 4 \),differentiate it with respect to time \( t \):\( v(t) = \frac{ds}{dt} = 6t^2 - 12t + 3 \).

Step 2: Find the acceleration

The acceleration is the first derivative of the velocity function. Differentiate the velocity function with respect to time \( t \):\( a(t) = \frac{dv}{dt} = 12t - 12 \).

Step 3: Find the time when acceleration is zero

Set the acceleration function equal to zero and solve for \( t \):\( 12t - 12 = 0 \).Solve for \( t \):\( t = 1 \).

Step 4: Find the directed distance at \( t = 1 \)

Substitute \( t = 1 \) into the position function to find the distance from the origin:\( s(1) = 2(1)^3 - 6(1)^2 + 3(1) - 4 \).Simplify the expression:\( s(1) = 2 - 6 + 3 - 4 = -5 \).So, the directed distance from the origin is -5 feet.

Step 5: Find the instantaneous velocity at \( t = 1 \)

Substitute \( t = 1 \) into the velocity function:\( v(1) = 6(1)^2 - 12(1) + 3 \).Simplify the expression:\( v(1) = 6 - 12 + 3 = -3 \).So, the instantaneous velocity is -3 ft/sec.

Key Concepts

Equation of Motion

The concept of the equation of motion is crucial in understanding how objects move. The equation given in the exercise, \[ s(t) = 2t^3 - 6t^2 + 3t - 4 \] represents the position of a particle as a function of time. Here, \( s \) is the position and \( t \) is the time. By analyzing this equation, we can derive important information such as velocity and acceleration. The equation of motion encapsulates how a particle's position changes over time, and it is key to solving many problems in physics and engineering.

Velocity

Velocity refers to the rate of change of position with respect to time. It's the first derivative of the position function. For the given exercise, the position function is \( s(t) = 2t^3 - 6t^2 + 3t - 4 \). To find the velocity, we take the first derivative of \( s(t) \) with respect to \( t \):\[ v(t) = \frac{ds(t)}{dt} = 6t^2 - 12t + 3 \]. This resulting function \( v(t) \) tells us how fast and in which direction the particle is moving at any given moment. When the velocity is positive, the particle is moving forward, and when it's negative, the particle is moving backward.

Acceleration

Acceleration describes how the velocity of an object changes over time. It's the first derivative of the velocity function. For the exercise at hand, we already have \( v(t) = 6t^2 - 12t + 3 \). To find the acceleration, we differentiate the velocity function with respect to time \( t \):\[ a(t) = \frac{dv(t)}{dt} = 12t - 12 \]. This function \( a(t) \) shows how the velocity changes at any instant \( t \). Setting this function equal to zero and solving for \( t \) gives us the time \( teq1 \) where acceleration is zero. This is significant because it points to moments when the particle's motion transitions in a notable way.

Derivatives

Derivatives are essential in calculus and provide a way to understand rates of change. In the context of motion, derivatives help us go from position functions to velocity and from velocity to acceleration. To find the velocity \( v(t) \) of a particle, we take the first derivative of the position function \( s(t) \):\[ v(t) = \frac{ds}{dt} \]. To find acceleration \( a(t) \), we then differentiate the velocity function:\[ a(t) = \frac{dv}{dt} \]. By mastering derivatives, we can analyze and understand the dynamic aspects of how objects move under different conditions. This allows us to predict future motion, calculate forces, and understand energy distributions in physical systems.