Problem 6 - Calculus Single Variable

Question

Describes the position of an object at time \(t .\) Calculate the instantaneous velocity at time \(c\). $$ p(t)=-5 t^{2}+2 \quad c=1 $$

Step-by-Step Solution

Answer

The instantaneous velocity at \(t = 1\) is \(-10\).

Step 1: Understand the Formula for Instantaneous Velocity

The instantaneous velocity of an object at a given time \(c\) can be found by computing the derivative of the position function \(p(t)\) with respect to \(t\), and evaluating it at \(t = c\). This means you need to find \(p'(t)\) and then evaluate \(p'(c)\).

Step 2: Find the Derivative of the Position Function

Given the position function \( p(t) = -5t^2 + 2 \), we need to differentiate it with respect to \(t\). The power rule states if \( f(t) = at^n \), then \( f'(t) = nat^{n-1} \). Apply this to \( -5t^2 \) to get \(-10t\), and the derivative of a constant \(2\) is \(0\). Thus, the derivative is \( p'(t) = -10t \).

Step 3: Evaluate the Derivative at t = c

Substitute \(t = 1\) into the derivative \( p'(t) = -10t \). This gives \( p'(1) = -10(1) = -10 \).

Step 4: State the Instantaneous Velocity

The instantaneous velocity at \(t = 1\) is \(-10\.\) This means that at time \(c = 1\), the object is moving with a velocity of \(-10\) units (could be meters or any other unit of length per unit of time) in its respective direction.

Key Concepts

Position Function

The position function is an essential concept when calculating motion in physics. It represents the location of an object as it changes over time. For this exercise, we have a position function given by \( p(t) = -5t^2 + 2 \). Here:

• \( t \) is the time variable.
• \(-5t^2 + 2\) is a quadratic equation, which will describe how the object's position varies at any given time \( t \).

The coefficients and terms in this equation provide valuable information about motion. For instance, the \(-5t^2\) term indicates acceleration affecting direction and speed, while "+2" could represent the object's initial position. Understanding the position function is crucial for determining any particulars about a moving object, such as velocity.

Derivative

A derivative in mathematics refers to the rate at which a function is changing at any given point. It is a fundamental tool in calculus and is used to calculate instantaneous rates of change, such as the velocity of a moving object. In our problem, finding the derivative of the position function \( p(t) = -5t^2 + 2 \) helps us determine the velocity function.For a function \( f(t) = -5t^2 \), the derivative \( f'(t) \) gives us the rate of change of position with respect to time.

To find the derivative of our specific position function, we apply calculus rules that simplify this process. This step prepares us to evaluate the motion at specific instants more effectively.

Power Rule

The power rule is a handy tool in calculus used for differentiating functions of the form \( f(t) = at^n \). For such functions, the derivative is \( f'(t) = nat^{n-1} \). This rule allows us to efficiently find derivatives without extensive computation. In our original exercise, to find the derivative of the position function \( p(t) = -5t^2 + 2 \), we apply the power rule to \(-5t^2\). Using the power rule step-by-step:

• The coefficient \( a = -5 \) is multiplied by the exponent \( n = 2 \), resulting in \(-10\).
• We decrease the exponent by one, moving from \( t^2 \) to \( t^1 \), leading to \(-10t\).

Therefore, the derived function is \( p'(t) = -10t \), which describes the velocity of the object at any moment \( t \).

Velocity Calculation

After determining the derivative of the position function, calculating the velocity is straightforward. Here, we want the instantaneous velocity at a specific time \( c = 1 \).With the derivative \( p'(t) = -10t \) from the previous steps, you substitute the given time \( t = 1 \) into this expression:Substitute:

• \( p'(1) = -10(1) = -10 \)

This result tells us that at the precise moment when \( t = 1 \), the instantaneous velocity of the object is \(-10\) (units per time). The negative sign in the result implies direction, indicating that the object moves in the opposite direction of the positive axis.

This calculation is essential, especially in real-time applications, as it provides insights into how fast and in which direction an object is traveling at an exact point in time.