Problem 27 - Precalculus: Mathematical for Calculus

Question

The displacement \(s\) (in meters) of a particle moving in a straight line is given by the equation of motion \(s=4 t^{3}+6 t+2,\) where \(t\) is measured in seconds. Find the velocity of the particle \(s\) at times \(t=a, t=1\) \(t=2, t=3\)

Step-by-Step Solution

Answer

The velocities are \(v(a) = 12a^2 + 6\), \(v(1) = 18\), \(v(2) = 54\), and \(v(3) = 114\).

Step 1: Understand the Velocity Concept

Velocity is the rate of change of displacement with respect to time. In calculus, it is found by taking the derivative of the displacement function concerning time.

Step 2: Differentiate the Displacement Function

To find the velocity function, differentiate the displacement equation, \(s(t) = 4t^3 + 6t + 2\), with respect to \(t\). The derivative, \(v(t)\), is given by: \(\frac{d}{dt}(4t^3 + 6t + 2) = 12t^2 + 6\). Thus, the velocity function is \(v(t) = 12t^2 + 6\).

Step 3: Calculate the Velocity at Different Times

Now, substitute the values \(t = a, 1, 2, 3\) into the derived velocity function to get the velocities. For\( t = a\), \(v(a) = 12a^2 + 6\). For \(t = 1\), \(v(1) = 12\cdot1^2 + 6 = 18\). For \(t = 2\), \(v(2) = 12\cdot2^2 + 6 = 54\). For \(t = 3\), \(v(3) = 12\cdot3^2 + 6 = 114\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It allows us to find the rate of change of a function concerning a variable. This rate of change is crucial for understanding how functions behave. In the context of motion, differentiation helps us understand how the position of an object changes over time by finding its velocity.
To perform differentiation, we use rules like the power rule, which is commonly applied in polynomials. For example, if we have the displacement function of a particle, such as \(s(t) = 4t^3 + 6t + 2\), differentiating it with respect to time \(t\) results in \(v(t) = 12t^2 + 6\). This result is obtained using the power rule: multiplying the power by the coefficient and then subtracting one from the power itself.
Differentiation is not only about applying formulas; it's about understanding how each operation translates to real-world changes. For example, when you differentiate displacement, you find how fast the displacement is changing, which is its velocity.

Velocity

Velocity describes how fast an object moves in a particular direction. Unlike speed, which is only a scalar and shows how fast something is moving, velocity is a vector because it also specifies the direction of movement. In our problem, understanding velocity involves taking the derivative of a displacement function.
For a displacement function \(s(t) = 4t^3 + 6t + 2\), the derivative yields the velocity function \(v(t) = 12t^2 + 6\). This process shows that velocity is contingent on the time \(t\) and changes as time progresses.
Once we have the velocity function, we can calculate the velocity of the particle at any given moment by simply substituting different values of \(t\). For instance, at \(t = 1\), the velocity is found to be \(v(1) = 18\). This gives us insights into how the particle's speed and direction evolve over time.

Equation of Motion

An equation of motion is a mathematical representation that describes the position of an object over time. In the context of the given problem, the equation of motion is \(s(t) = 4t^3 + 6t + 2\), which provides the displacement of a particle in relation to time.
Understanding the equation of motion allows us to predict and analyze the past, present, and future state of a particle's position. Differentiating this equation gives mathematical insights into velocity and acceleration, defining the trajectory of motion.

• Displacement: The actual function \(s(t) = 4t^3 + 6t + 2\) shows where the particle is at any moment \(t\).
• Velocity: Derived by differentiating the displacement equation, indicating how fast and in what direction.
• Acceleration: This would be the next derivative, telling how the velocity changes over time.

From the displacement equation, taking the derivative allows us to transition from understanding static positions to dynamic movements and eventually to change in velocity (acceleration). This forms the basis for analyzing and predicting motion scientifically.