Problem 46 - Calculus Single Variable

Question

The position of a moving body is described by \(p(t)=\) \(a t^{2}+b t+c .\) If \(p(0)=3,\) then what is \(c ?\) If \(p^{\prime}(0)=6,\) then what is \(b\) ? If \(p^{\prime \prime}(0)=-5,\) then what is \(a ?\) Where is the body at time \(t=6 ?\)

Step-by-Step Solution

Answer

At \(t=6\), the body is at position \(-51\).

Step 1: Understanding the equation at time t=0

Given the position function \(p(t) = at^2 + bt + c\), and \(p(0) = 3\), substituting \(t = 0\) gives \(c = 3\). Hence, the constant \(c = 3\).

Step 2: Finding b using the velocity at t=0

The velocity is given by the first derivative \(p'(t) = 2at + b\). With \(p'(0) = 6\), substituting \(t = 0\) gives \(b = 6\).

Step 3: Determining a using the acceleration at t=0

The acceleration is given by the second derivative \(p''(t) = 2a\). Since \(p''(0) = -5\), we solve \(2a = -5\), thus \(a = -\frac{5}{2}\).

Step 4: Evaluating the position at t=6

The position equation is \(p(t) = at^2 + bt + c\) with \(a = -\frac{5}{2}\), \(b = 6\), and \(c = 3\). Substitute \(t = 6\) to find \(p(6) = -\frac{5}{2}(6)^2 + 6(6) + 3\).

Step 5: Calculating the final position

Substitute values to simplify: \(p(6) = -\frac{5}{2}(36) + 36 + 3 = -90 + 36 + 3 = -51\). Therefore, the position at \(t = 6\) is \(-51\).

Key Concepts

First Derivative

The concept of a derivative is fundamental in calculus, allowing us to determine the rate at which a function changes. Specifically, the first derivative of a function provides us with the rate of change of the function with respect to its variable. In kinematics, this first derivative is critical as it represents the velocity of a moving object.
For a position function like \(p(t) = at^2 + bt + c\), the first derivative, \(p'(t) = 2at + b\), describes how fast the position is changing over time.

• At any given time \(t\), \(p'(t)\) tells us the instantaneous velocity of the body.
• In the given problem, when \(p'(0) = 6\), we substitute \(t = 0\) into \(p'(t)\) to find \(b = 6\).

Understanding the first derivative is crucial in analyzing the motion's dynamics, especially when calculating how quickly a body is moving at a specific moment.

Second Derivative

When we take the derivative of a function's first derivative, we get the second derivative. This second derivative provides even deeper insights into the behavior of a function by telling us about the acceleration. In kinematics, acceleration is the rate of change of velocity with respect to time.
In our case, the second derivative of the position function \(p(t) = at^2 + bt + c\) is given by \(p''(t) = 2a\).

• This simple linear function demonstrates that acceleration is constant in quadratic position functions.
• For the problem at hand, with \(p''(0) = -5\), we substitute into the equation to solve for \(a\), finding that \(a = -\frac{5}{2}\).

This constant acceleration could imply that forces acting on the body are constant as well, influencing the motion in a uniform manner.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two, characterized in the standard form as \(ax^2 + bx + c = 0\). These equations are paramount in modeling various real-life scenarios, including the path of moving objects in physics.
When analyzing motion, a quadratic equation often encapsulates the position function of an object. The coefficients \(a\), \(b\), and \(c\) provide specific information:

• \(a\) determines the concavity of the parabola representing the path.
• \(b\) influences the direction and symmetry.
• \(c\) indicates the initial position when time is zero.

By solving quadratic equations, one can predict future positions and motion behaviors, essential for accurate kinematic analysis.

Kinematics

Kinematics is the branch of physics that studies motion without considering the forces that cause it. It involves analyzing patterns of movement to understand how an object moves over time.
A position function, such as \(p(t) = at^2 + bt + c\), is a classic tool in kinematics used to describe an object's location at any given time \(t\). Kinematics focuses on several key characteristics of motion:

• Position, represented by the quadratic equation.
• Velocity, derived from the first derivative.
• Acceleration, obtained from the second derivative.

In this specific exercise, we see how kinematics allows us to calculate the body’s position at \(t = 6\) using the previously determined values of \(a\), \(b\), and \(c\). By plugging in \(t = 6\) to the position function, we find that the body is located at position \(-51\). Understanding these concepts is fundamental for deeper applications in physics and engineering.