Problem 23 - The Calculus with Analytic Geometry

Question

An object is moving along a straight line according to the equation of motion \(s=\sqrt{4 t^{2}+3}\), with \(t \geq 0\). Find the values of \(t\) for which the measure of the instantaneous velocity is (a) \(0 ;\) (b) \(1 ;\) (c) 2 .

Step-by-Step Solution

Answer

For (a) t = 0; for (b) t = \frac{1}{2}; for (c) no real solution.

Step 1: Understand the given equation

The equation of motion is given by \[s = \sqrt{4t^{2} + 3}\].

Step 2: Find the velocity function

The instantaneous velocity, v(t), is the derivative of the position function, s(t), with respect to time, t. First, find the derivative of \[s = \sqrt{4 t^{2}+3}\].

Step 3: Use the chain rule for differentiation

The chain rule is needed to differentiate \[s(t) = \sqrt{4 t^{2}+3}\].Let \[u = 4 t^{2} + 3\]. Then \[s = \sqrt{u}\] and rewrite \[v(t) = \frac{ds}{dt} = \frac{d}{dt} (u^{1/2}) = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dt}\].

Step 4: Differentiate u with respect to t

Need to differentiate \[u = 4 t^{2} + 3\] with respect to t. This gives \[\frac{du}{dt} = 8 t\].

Step 5: Find the final expression for v(t)

Combine the results from earlier steps: \[v(t) = \frac{1}{2} (4 t^{2}+3)^{-1/2} \cdot 8 t\].Simplify to get: \[v(t) = \frac{4 t}{\sqrt{4 t^{2} + 3}}\].

Step 6: Set up the equation for part (a)

Set the expression for v(t) equal to 0 to solve for t. This gives \[\frac{4 t}{\sqrt{4 t^{2} + 3}} = 0\].

Step 7: Solve for t in part (a)

The numerator of the fraction must be 0 for the fraction to equal 0. Therefore, \[4 t = 0\].Solve to get \[t = 0\].

Step 8: Set up the equation for part (b)

Set the expression for v(t) equal to 1 to solve for t. This gives \[\frac{4 t}{\sqrt{4 t^{2} + 3}} = 1\].

Step 9: Solve for t in part (b)

Square both sides to remove the square root: \[\left(\frac{4 t}{\sqrt{4 t^{2} + 3}}\right)^{2} = 1^{2}\].Simplify: \[\frac{16 t^{2}}{4 t^{2} + 3} = 1\].Cross-multiply: \[16 t^{2} = 4 t^{2} + 3\].Rearrange: \[12 t^{2} = 3\].Solve for t: \[t^{2} = \frac{1}{4}\], giving \[t = \frac{1}{2}\].

Step 10: Set up the equation for part (c)

Set the expression for v(t) equal to 2 to solve for t. This gives \[\frac{4 t}{\sqrt{4 t^{2} + 3}} = 2\].

Step 11: Solve for t in part (c)

Square both sides to remove the square root: \[\left(\frac{4 t}{\sqrt{4 t^{2} + 3}}\right)^{2} = 2^{2}\].Simplify: \[\frac{16 t^{2}}{4 t^{2} + 3} = 4\].Cross-multiply: \[16 t^{2} = 16 t^{2} + 12\].Rearrange and solve: \[-12 = 0\].This is a contradiction, so no real solution for t.

Key Concepts

Equation of Motion

The equation of motion is a fundamental concept that describes how the position of an object changes over time. In our case, the equation of motion is given by: i.e. The variable represents time, and represents the position of the object at any given time . Understanding this equation allows us to determine crucial aspects of the object's movement.

Chain Rule Differentiation

To find the instantaneous velocity, we must differentiate the position function with respect to time. Here, the position function is . Since this function is a composition of two simpler functions and , we use the chain rule for differentiation. The chain rule states that if we have a composed function , then its derivative is given by Using this rule: The chain rule provides an efficient way to handle functions inside other functions, which is quite common in physics and engineering problems.

Solving Equations

Once we have derived the velocity function, , we solve it for specific values to determine when the instantaneous velocity is a certain value. This process involves setting up an equation and solving for the variable . For instance: For : , simplifying this equation confirms the time at which the velocity meets the specified condition. Understanding how to solve these equations is essential as it applies to a wide range of problems beyond just physics, including various fields of engineering and mathematics.