Problem 99 - University Calculus: Early Transcendentals

Question

The position of a particle moving along a coordinate line is \(s=\sqrt{1+4 t},\) with \(s\) in meters and \(t\) in seconds. Find the particle's velocity and acceleration at \(t=6\) sec.

Step-by-Step Solution

Answer

At \( t = 6 \) seconds: velocity is \( \frac{2}{5} \) m/s, acceleration is \( -\frac{4}{125} \) m/s².

Step 1: Find the Velocity Function

The velocity of a particle is the derivative of the position function with respect to time. Thus, if \( s(t) = \sqrt{1 + 4t} \), then the velocity function \( v(t) \) is the derivative of \( s(t) \) with respect to time \( t \). To find this derivative, use the chain rule. Rewrite the position function as \( s(t) = (1 + 4t)^{1/2} \).Using the chain rule, the derivative is:\[ v(t) = \frac{d}{dt}(1 + 4t)^{1/2} = \frac{1}{2}(1 + 4t)^{-1/2} \cdot 4 = \frac{2}{\sqrt{1 + 4t}} \].

Step 2: Evaluate Velocity at t = 6 Seconds

Now that we have the velocity function \( v(t) = \frac{2}{\sqrt{1 + 4t}} \), we need to find the velocity at \( t = 6 \) seconds. Substitute \( t = 6 \) into the velocity function:\[ v(6) = \frac{2}{\sqrt{1 + 4 \cdot 6}} = \frac{2}{\sqrt{1 + 24}} = \frac{2}{5} \]. Therefore, the velocity at \( t = 6 \) seconds is \( \frac{2}{5} \) meters per second.

Step 3: Find the Acceleration Function

The acceleration of a particle is the derivative of the velocity function with respect to time. We already found that the velocity function is \( v(t) = \frac{2}{\sqrt{1 + 4t}} \).To find its derivative, apply the chain and quotient rules:Setting \( u(t) = \sqrt{1 + 4t} = (1 + 4t)^{1/2} \), the derivative is:\[ a(t) = \frac{d}{dt}(\frac{2}{u}) = -\frac{2}{u^2} \cdot \frac{d}{dt}(u) \].Given \( u = (1 + 4t)^{1/2} \) and \( \frac{d}{dt}u = \frac{2}{\sqrt{1 + 4t}} \), calculate the derivative:\[ a(t) = -\frac{2 \cdot 2}{(1 + 4t)^{3/2}} \cdot 2 = -\frac{4}{(1 + 4t)^{3/2}} \].

Step 4: Evaluate Acceleration at t = 6 Seconds

Now that we have the acceleration function \( a(t) = -\frac{4}{(1 + 4t)^{3/2}} \), find the acceleration at \( t = 6 \) seconds.Substitute \( t = 6 \) into the acceleration function:\[ a(6) = -\frac{4}{(1 + 4 \cdot 6)^{3/2}} = -\frac{4}{5^3} = -\frac{4}{125} \].Thus, the acceleration at \( t = 6 \) seconds is \( -\frac{4}{125} \) meters per second squared.

Key Concepts

Derivatives

In calculus, a derivative represents the rate at which a quantity changes. It is a fundamental concept that helps us understand how functions change over time.
When dealing with motion, such as the movement of a particle along a line, the derivative of the position function concerning time gives us the velocity function. This tells us how fast the particle is moving and in which direction.

• Position Function (s): This gives the position of the particle at any time, expressed as a function of time, like \( s(t) = \sqrt{1 + 4t} \) in our example.
• Velocity Function (v): To find it, take the derivative of the position function. This involves using differentiation rules like the chain rule or the quotient rule.

For a position function \( s(t) \), the velocity is \( v(t) = \frac{d}{dt}s(t) \). Using differentiation techniques allows us to explore functions with complex terms or those involving powers or roots.

In the given solution, we transitioned the function from \( s(t) = \sqrt{1 + 4t} \) to \( s(t) = (1 + 4t)^{1/2} \) to simplify the differentiation process, applying the chain rule to obtain the velocity function. Understanding derivatives is key to solving problems involving changing rates, making them useful not merely in mathematics but in numerous real-world applications.

Velocity

Velocity offers insight into the speed and direction of a moving object. It is a vector quantity combining both magnitude and direction, unlike speed which only measures magnitude.
In the context of our problem, once we determined the velocity function using derivatives, it is vital to evaluate it at specific moments to understand the particle's motion at those times.

• Evaluating velocity: Substitute a specific time into the velocity function to gauge the particle's speed at that instant.

For instance, in the given solution, evaluating at \( t = 6 \) seconds showed us the particle's velocity as \( \frac{2}{5} \) meters per second. This not only tells how fast the particle is moving but also provides direction information (positive or negative values indicating forward or backward motion respectively).

Understanding the velocity helps in predicting future positions and planning motion adjustments if needed, making it essential in physics and engineering.

Acceleration

Acceleration measures the rate at which velocity changes over time. It is a crucial concept in calculus that provides information on how an object's speed is increasing or decreasing.
Calculating acceleration begins with the velocity function; taking its derivative yields the acceleration function.

• Finding Acceleration: The acceleration function is the derivative of the velocity function concerning time.
• Interpreting Acceleration: Positive values indicate speeding up, while negative values suggest slowing down.

In the solved example, we derived the acceleration function from the velocity function \( v(t) = \frac{2}{\sqrt{1 + 4t}} \). By applying the quotient and chain rules, we found the acceleration function \( a(t) = -\frac{4}{(1 + 4t)^{3/2}} \). Evaluating this at \( t = 6 \) seconds gave us an acceleration of \( -\frac{4}{125} \) meters per second squared, indicating a deceleration or slowing down of the particle.

Grasping acceleration is important in many fields, especially those involving any form of dynamics like mechanical or aerospace engineering. It's about understanding changes in motion and making informed predictions or safety assessments.