Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(s=\frac{1}{10}\left(t^{4}-14 t^{3}+60 t^{2}\right)\), find the velocity of the moving object when its acceleration is zero.

Step-by-Step Solution

Answer

The object has velocity 50 units/s at \(t=5\) and 1.6 units/s at \(t=2\) when acceleration is zero.

Step 1: Understand the Problem

We need to find the velocity when acceleration is zero. This means that the second derivative of the position function is zero, and the velocity is the first derivative of the position function.

Step 2: Write Down the Position Function

The position function is given as \(s = \frac{1}{10} (t^4 - 14t^3 + 60t^2)\).

Step 3: Find the Velocity Function

The velocity function \(v(t)\) is the first derivative of the position function with respect to time \(t\). \[ v(t) = \frac{d}{dt}\left(\frac{1}{10} (t^4 - 14t^3 + 60t^2) \right) = \frac{1}{10} (4t^3 - 42t^2 + 120t) \].

Step 4: Find the Acceleration Function

The acceleration function \(a(t)\) is the derivative of the velocity function. \[ a(t) = \frac{d}{dt}(\frac{1}{10} (4t^3 - 42t^2 + 120t)) = \frac{1}{10} (12t^2 - 84t + 120) \].

Step 5: Set the Acceleration to Zero

To find when the acceleration is zero, solve the equation: \[ \frac{1}{10} (12t^2 - 84t + 120) = 0 \].This simplifies to the quadratic equation: \[ 12t^2 - 84t + 120 = 0 \].

Step 6: Solve the Quadratic Equation

First, simplify the equation by dividing by 12: \[ t^2 - 7t + 10 = 0 \].Now, use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1\), \(b=-7\), \(c=10\). \[ t = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm 3}{2} \].This gives solutions \(t = 5\) and \(t = 2\).

Step 7: Find the Velocity at the Identified Times

Substitute \(t = 5\) and \(t = 2\) into \(v(t) = \frac{1}{10} (4t^3 - 42t^2 + 120t) \) to find the velocities:For \(t = 5\):\[ v(5) = \frac{1}{10} (4(5)^3 - 42(5)^2 + 120(5)) = \frac{1}{10} (500) \] \[ = 50 \].For \(t = 2\):\[ v(2) = \frac{1}{10} (4(2)^3 - 42(2)^2 + 120(2)) = \frac{1}{10} (16) \]\[ = 1.6 \].

Key Concepts

Velocity

Velocity is a key concept in physics and calculus, representing the rate at which an object changes its position. It is a vector quantity, which means it has both magnitude and direction.
In mathematics, velocity is derived from the position function, which describes an object's location over time. It's calculated as the first derivative of the position function with respect to time. This tells us how fast and in what direction the position is changing at any given point in time.

• For instance, if the position function is given as \(s(t)\), the velocity \(v(t)\) is \( \frac{ds}{dt} \).
• This derivative measures the instantaneous rate of change of the position.

Understanding velocity is crucial for solving problems involving motion, like calculating how far a car has traveled over a period of time or determining its speed.

Acceleration

Acceleration describes how the velocity of an object changes with time. It is also a vector quantity and looks at the rate of change of velocity.
Simply put, acceleration can tell us if an object is speeding up, slowing down, or changing direction. Like velocity, acceleration is found by taking a derivative.

• The acceleration \(a(t)\) is the derivative of the velocity function \(v(t)\).
• If \(v(t)\) is the first derivative of the position function \(s(t)\), then \(a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} \).

When acceleration equals zero, it means that the velocity is constant during that interval. This is an important concept, especially in physics, as it can simplify calculations in motion problems.

Derivative

The derivative is a fundamental concept in calculus that describes how a function changes at any point. It provides the rate at which one quantity changes with respect to another.
For motion, derivatives help us move from position to velocity and then to acceleration:

• The first derivative of a position function with respect to time is velocity, \(v(t) = \frac{ds}{dt} \).
• The second derivative is acceleration, \(a(t) = \frac{d^2s}{dt^2} \).

Derivatives can be thought of as the "slope" of a function at a point. In the context of motion, derivatives offer precise mathematical descriptions of the dynamic changes in a moving system.

Quadratic Equation

Quadratic equations appear frequently in calculus, especially in problems involving motion.
A quadratic equation typically has the form \(ax^2 + bx + c = 0\), and can have two, one, or no real solutions. These equations are solved using either factoring, completing the square, or the quadratic formula.

• The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) provides a straightforward method to find the roots of the equation.
• In our specific problem, setting the acceleration to zero led to a quadratic equation in terms of time \(t\), \(12t^2 - 84t + 120 = 0\).
• Solving this helped identify the exact times when the velocity measurement was needed.

Understanding how to work with quadratic equations is essential because they help us express and solve real-world problems where relationships between variables are not linear.