Question

If \(s=\frac{1}{2} t^{4}-5 t^{3}+12 t^{2}\), find the velocity of the moving object when its acceleration is zero.

Short answer

The possible velocities are 11 or -16.

Step-by-step solution

Find the velocity function

Velocity is the first derivative of the position function with respect to time. The given position function is \( s = \frac{1}{2}t^4 - 5t^3 + 12t^2 \). To find the velocity function \( v(t) \), we differentiate \( s \) with respect to \( t \): \[ v(t) = \frac{d}{dt}\left(\frac{1}{2}t^4 - 5t^3 + 12t^2\right) = 2t^3 - 15t^2 + 24t \].

Find the acceleration function

Acceleration is the first derivative of the velocity function or the second derivative of the position function. Differentiate the velocity function \( v(t) = 2t^3 - 15t^2 + 24t \) with respect to \( t \):\[ a(t) = \frac{d}{dt}(2t^3 - 15t^2 + 24t) = 6t^2 - 30t + 24 \].

Set the acceleration to zero

To find when the acceleration is zero, set the derived acceleration function equal to zero and solve for \( t \):\[ 6t^2 - 30t + 24 = 0 \]. Divide by 6 to simplify:\[ t^2 - 5t + 4 = 0 \]. Factor the quadratic:\[ (t-1)(t-4) = 0 \]. Thus, the solutions are \( t = 1 \) and \( t = 4 \).

Calculate the velocity at \( t = 1 \) and \( t = 4 \)

Substitute \( t = 1 \) and \( t = 4 \) into the velocity function to find the velocity at these times:For \( t = 1 \):\[ v(1) = 2(1)^3 - 15(1)^2 + 24(1) = 2 - 15 + 24 = 11 \].For \( t = 4 \):\[ v(4) = 2(4)^3 - 15(4)^2 + 24(4) = 128 - 240 + 96 = -16 \].

Conclude with the correct velocity

With acceleration equal to zero, the velocities were calculated as \( v(1) = 11 \) and \( v(4) = -16 \). The velocity of the moving object could be either 11 or -16, based on the specific context required, such as the given time interval or constraints.

Key concepts

Derivatives

Derivatives are a fundamental concept in calculus and serve as the mathematical tool for determining rates of change. When we take the derivative of a function with respect to a variable, we are essentially calculating how the function's output changes as its input changes. In the context of motion, derivatives allow us to move from position functions to velocity and acceleration functions.

For example, if you have a position function that describes how an object's position changes over time, taking the first derivative of that function with respect to time gives you the velocity function. This velocity function then tells you how quickly and in what direction the object's position is changing at any given point in time. Similarly, taking the derivative of the velocity function gives you the acceleration function, which describes the rate of change of velocity.

• Derivatives help in finding instantaneous rates of change.
• They are used in various applications, from physics to economics.
• Understanding derivatives is crucial for studying dynamic systems that change over time.

Velocity

Velocity is a vector quantity that indicates both the speed and direction of an object's movement. In calculus, velocity is determined by the first derivative of the position function with respect to time. This reveals the object's rate of change of position and is expressed as a function of time.

For the position function given by \( s = \frac{1}{2}t^4 - 5t^3 + 12t^2 \), the velocity function is derived as \( v(t) = 2t^3 - 15t^2 + 24t \) through differentiation. This function represents how the speed of the object changes over time. To find specific velocities at particular times, substitute the time values into this velocity function. For instance, when the acceleration is zero, the problem requires checking the velocity at those crucial points in time, which could lead to either a positive increase in speed or a decrease.

Acceleration

Acceleration is another vector quantity and represents the rate of change of velocity over time. In calculus, this is found by taking the derivative of the velocity function, making acceleration the second derivative of the position function with respect to time.

Given the velocity function \( v(t) = 2t^3 - 15t^2 + 24t \), the acceleration function is derived as \( a(t) = 6t^2 - 30t + 24 \). This expression helps determine when an object is speeding up or slowing down.

To find when the acceleration is zero, set \( a(t) = 0 \) and solve the resulting quadratic equation. Finding this zero-point can indicate a change in the behavior of the object's velocity, such as shifting from accelerating to decelerating, or achieving steady motion momentarily.

Quadratic Equations

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Quadratic equations often appear in physics problems involving motion, such as when setting the acceleration function to zero to find critical points.

In the context of the exercise, the acceleration function simplifies to \( t^2 - 5t + 4 = 0 \) when seeking where acceleration equals zero. This is a standard quadratic equation, and it can be solved by factoring method, completing the square, or using the quadratic formula.

• Solving these equations is essential for determining key moments in motion problems, like when an object changes direction.
• Quadratics can have two solutions, no real solutions, or one repeated solution.
• Understanding how to solve these is vital in predicting outcomes in many applied mathematics scenarios.