Problem 38 - Calculus Volume 1

Question

IT] For the following position functions \(y=s(t), \quad\) an object is moving along a straight line, where \(t\) is in seconds and \(s\) is in meters. Find a. the simplified expression for the average velocity from \(t=2\) to \(t=2+h\) b. the average velocity between \(t=2\) and \(t=2+h, \quad\) where \((\mathrm{i}) h=0.1, \quad\) (ii) \(h=0.01\) (iii) \(h=0.001,\) and (iv) \(h=0.0001 ;\) and c. use the answer from a. to estimate the instantaneous velocity at \(t=2\) second. $$ s(t)=\frac{16}{t^{2}}-\frac{4}{t} $$

Step-by-Step Solution

Answer

The simplified expression for average velocity is \( v_{avg} = \frac{\left(\frac{16}{(2+h)^{2}} - \frac{4}{(2+h)}\right) - 2}{h} \), and the instant velocity at \(t=2\) is \(-1\) m/s.

Step 1: Simplify the Expression for Average Velocity

The formula for average velocity between time \(t=2\) and \(t=2+h\) is given by:\[ v_{avg} = \frac{s(2+h) - s(2)}{h} \]First, we calculate \(s(2+h)\):\[ s(t) = \frac{16}{t^{2}} - \frac{4}{t} \]Substitute \(t = 2+h\):\[ s(2+h) = \frac{16}{(2+h)^{2}} - \frac{4}{(2+h)} \]Calculate \(s(2)\):\[ s(2) = \frac{16}{2^2} - \frac{4}{2} = 4 - 2 = 2 \]Finally, plug these into the formula for \(v_{avg}\):\[ v_{avg} = \frac{\left(\frac{16}{(2+h)^{2}} - \frac{4}{(2+h)}\right) - 2}{h} \]

Step 2: Evaluate Limits for Average Velocity Values

To evaluate the average velocities for different values of \(h\), substitute each value into the expression derived:(i) For \(h = 0.1\):\[ v_{avg} = \frac{\left(\frac{16}{2.1^{2}} - \frac{4}{2.1}\right) - 2}{0.1} \]after calculating, the approximate value is: \(v_{avg} \approx -0.952 \)(ii) For \(h = 0.01\):\[ v_{avg} = \frac{\left(\frac{16}{2.01^{2}} - \frac{4}{2.01}\right) - 2}{0.01} \]after calculating, the approximate value is: \(v_{avg} \approx -0.990 \)(iii) For \(h = 0.001\):\[ v_{avg} = \frac{\left(\frac{16}{2.001^{2}} - \frac{4}{2.001}\right) - 2}{0.001} \]after calculating, the approximate value is: \(v_{avg} \approx -0.999 \)(iv) For \(h = 0.0001\):\[ v_{avg} = \frac{\left(\frac{16}{2.0001^{2}} - \frac{4}{2.0001}\right) - 2}{0.0001} \]after calculating, the approximate value is: \(v_{avg} \approx -0.9999 \)

Step 3: Estimate Instantaneous Velocity

Instantaneous velocity at \(t=2\) is defined as the limit of the average velocity as \(h\) approaches zero. From the trend observed in Step 2, as \(h\) decreases, the average velocity approaches \(-1\).Thus, the instantaneous velocity at \(t=2\) is:\[ v_{inst} = \lim_{h \to 0} v_{avg} = -1 \text{ m/s} \]

Key Concepts

Average Velocity

When we talk about average velocity, we're referring to the overall change in position over a particular time interval. It’s a way to see how fast something is moving, ignoring any changes in speed along the way.
The formula to find average velocity is:

• \[ v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \]
  

In this example, since we're considering a very small timeframe from \(t = 2\) to \(t = 2 + h\), the expression becomes:

• \[ v_{avg} = \frac{s(2+h) - s(2)}{h} \]
  

This form, with \(h\), allows us to approach instantaneous velocity, which is central in calculus studies. Understanding this helps us see how close the average velocity gets to what’s happening at just one moment.

Limit Evaluation

Limits are a fundamental concept in calculus that provide a way to evaluate an expression as it gets infinitely close to a precise point. They allow us to handle scenarios where straightforward calculations would otherwise be challenging.
In the context of velocity, limits help us bridge the gap between average and instantaneous velocities. If we calculate the average velocity at very small increments (like our given values \(h = 0.1, 0.01, 0.001, 0.0001\)), we can see it approaching a specific value.
By taking the limit of the average velocity as \(h\) approaches zero, we essentially capture the velocity of the object at an exact time point. This is crucial as it helps to establish the instantaneous velocity, which is key to understanding motion dynamics at a granular level.

Position Function

A position function tells us where an object is at any given time. It’s like having a roadmap of an object's journey on a straight line, providing insight into its movement over time.
The specific position function given in the problem is:

• \[ s(t) = \frac{16}{t^2} - \frac{4}{t} \]
  

This formula calculates the position \(s\) of an object in meters, based on time \(t\) in seconds. For instance, when \(t = 2\), plugging this in gives us:

• \[ s(2) = \frac{16}{4} - \frac{4}{2} = 2 \] meters
  

Understanding the position function is the first step to calculating velocity, as changes in position over time (given by \(s(2+h)\) and \(s(2)\)) will directly affect velocity calculations.

Velocity Calculation

Velocity tells us how fast and in which direction something moves. Calculating velocity can be simple or complex depending on whether you're looking at average or instantaneous speed.
To calculate instantaneous velocity, we first start with average velocity over progressively smaller time intervals. The step-size \(h\) used in calculations affects how close our average velocity estimation is to instantaneous velocity:

• For larger \(h\), the rate might not be as accurate.
• As \(h\) approaches zero, our average velocity gets closer to the true instantaneous velocity.

In our example:

• \[ v_{inst} = \lim_{h \to 0} \frac{s(2+h) - s(2)}{h} = -1 \text{ m/s} \]
  

This result is crucial, as it represents the speed of the object exactly at \(t = 2\), with no consideration of previous or future movement, offering a snapshot of the object’s motion.