Problem 41 - Calculus Single Variable

Question

The instantaneous rate of change of velocity is acceleration. For the position function \(p(t)=t^{3},\) what is the acceleration at time \(t=1 ?\)

Step-by-Step Solution

Answer

The acceleration at time \(t=1\) is 6 units/s².

Step 1: Find the Velocity Function

To determine acceleration, we first need the velocity function. The velocity is the first derivative of the position function with respect to time. Given the position function \(p(t) = t^3\), the velocity function is \(v(t) = \frac{dp}{dt} = 3t^2\).

Step 2: Find the Acceleration Function

The acceleration is the derivative of the velocity function. Differentiate \(v(t) = 3t^2\) with respect to \(t\) to find the acceleration function: \(a(t) = \frac{dv}{dt} = 6t\).

Step 3: Calculate Acceleration at the Specific Time

Substitute \(t = 1\) into the acceleration function to find the acceleration at that time. So, \(a(1) = 6 \times 1 = 6\).

Key Concepts

Understanding Derivatives in Calculus

Derivatives form the backbone of calculus, dealing with how things change. Imagine a curve on a graph. The derivative at a specific point tells you the slope of the tangent to that curve at that point. It speaks of how the curve inclines or declines.

• Instantaneous Rate of Change: Derivatives measure this. How fast is something changing now?
• Notation: If you see \( \frac{dy}{dx} \), this denotes the derivative of \( y \) with respect to \( x \).

Taking derivatives involves rules and techniques like the power rule, product rule, and chain rule, to name a few. In our exercise, we first differentiate the position function to find velocity.

Velocity Explained

Velocity is not just speed; it has direction. In calculus terms, velocity is the derivative of the position function. With it, you understand how fast an object changes its position at any given time.

• Obtaining Velocity: It requires differentiating the position function, \( p(t) \).
• Example: With \( p(t) = t^3 \), the velocity \( v(t) \) is \( \frac{dp}{dt} = 3t^2 \).

This gives an equation that describes how speed increases or decreases over time. Calculating velocity is crucial in physics and engineering for designing motion systems.

Delving into Acceleration

Acceleration tells us how the speed (or velocity) changes. While velocity describes movement, acceleration changes the movement dynamics.

• Acceleration Calculation: It's the first derivative of velocity, the second of the position.
• From Example: For \( v(t) = 3t^2 \), acceleration \( a(t) \) is \( \frac{dv}{dt} = 6t \).

Once the acceleration function is found, plugging in specific times can show the acceleration of the object precisely at those moments. This helps in designing systems and predicting future motion accurately.

Grasping the Position Function

The position function \( p(t) \) tells you where something is at time \( t \). It forms the basis for finding velocity and acceleration.

• Definition: This function sets the stage, showing paths or trajectories.
• Our Example: With \( p(t) = t^3 \), you know that position changes cubically with time.

Understanding the position function is crucial because, without it, we cannot begin to explore effortlessly the related ideas of speed and acceleration. It provides the context, the fundamental layer to comprehend the dynamics of motion.