Problem 35 - Calculus Volume 1

Question

For the following position functions \(y=s(t),\) 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 \((\) i) \(h=0.1\) (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. \(s(t)=\frac{1}{3} t+5\)

Step-by-Step Solution

Answer

The instantaneous velocity at \(t=2\) is \(\frac{1}{3}\) m/s.

Step 1: Simplify the Expression for Average Velocity

To find the average velocity from time \(t=2\) to \(t=2+h\), use the formula: \[ \text{Average Velocity} = \frac{s(2+h) - s(2)}{h}. \] Substitute \(s(t) = \frac{1}{3}t + 5\) into the equation:\[ s(2+h) = \frac{1}{3}(2+h) + 5 = \frac{1}{3} \cdot 2 + \frac{1}{3} \cdot h + 5 = \frac{2}{3} + \frac{h}{3} + 5. \] Hence, the expression for \(s(2+h) - s(2)\) is: \[ \left( \frac{2}{3} + \frac{h}{3} + 5 \right) - \left( \frac{2}{3} + 5 \right) = \frac{h}{3}. \] The simplified expression for \( \text{Average Velocity} \) is \( \frac{h/3}{h} = \frac{1}{3}. \) Hence, \( \text{Average Velocity} = \frac{1}{3}.\)

Step 2: Calculate Average Velocity for Different Values of h

Now, calculate the average velocity for the specified values of \(h\): (i) For \(h = 0.1\), the average velocity is \( \frac{1}{3} \) m/s.(ii) For \(h = 0.01\), the average velocity is \( \frac{1}{3} \) m/s.(iii) For \(h = 0.001\), the average velocity is \( \frac{1}{3} \) m/s.(iv) For \(h = 0.0001\), the average velocity still remains \( \frac{1}{3} \) m/s. In each case, the average velocity simplifies to the constant value of \( \frac{1}{3} \) m/s.

Step 3: Estimate Instantaneous Velocity at t=2

The instantaneous velocity at \(t=2\) is the limit of the average velocity as \(h\) approaches zero. Since the expression for average velocity simplifies to the constant \( \frac{1}{3} \) no matter the value of \(h\), the instantaneous velocity at \(t=2\) is also \( \frac{1}{3} \) m/s.

Key Concepts

Instantaneous Velocity

Imagine a car traveling along a road. If we want to know how fast it's going at an exact moment, that's its instantaneous velocity. In calculus, instantaneous velocity is the velocity of an object at a specific instant of time. It's a bit like the speedometer of your car showing 60 mph at a particular second. To find it, we often have to think about limits.We calculate instantaneous velocity by taking the limit of the average velocity as time intervals become extremely small. This means looking at how quickly a position changes over an almost zero length of time. Mathematically, if we consider a position function \( s(t) \), the instantaneous velocity at a specific time \( t \) is obtained by evaluating:\[ \lim_{h \to 0} \frac{s(t+h) - s(t)}{h} \]In our problem, we found the average velocity using a specific approach and determined it to be \( \frac{1}{3} \) m/s for every \( h \) given. Thus, the instantaneous velocity at \( t=2 \) is also \( \frac{1}{3} \) m/s. The use of limits in calculus helps to pinpoint these precise measures of speed, offering a powerful tool in understanding motion.

Position Function

The position function, commonly represented as \( s(t) \), is a mathematical formula that shows how an object's position changes with time. It's like a GPS tracker telling you where an object is as you keep moving it forward in time.In any motion-related scenario, understanding the position function is key. It tells us the exact position of an object at any given time \( t \). In the original exercise, we have: \[ s(t) = \frac{1}{3}t + 5 \]This particular equation signifies a straight-line motion where the object's position increases steadily by \( \frac{1}{3} \) meters for every additional second. The \( +5 \) indicates its initial position. Whether dealing with a simple or complex function, the position function provides the necessary connection from time to the spatial position, crucial for further calculations like velocity.

Calculus Problems

Dealing with calculus problems like finding velocities can feel daunting at first, but breaking them down makes them simpler. Calculus is about change, and it frequently addresses how rates and accumulations shift over time.Here’s how you can tackle calculus problems:

• **Understand the Function**: Make sure to interpret the function given, like \( s(t) = \frac{1}{3}t + 5 \), which describes how a position varies.
• **Set Up the Problem**: Clearly establish what you're asked to find. Are we looking for the average speed, an instantaneous value, or something different?
• **Use Limits and Derivatives**: When calculating velocities or accelerations, limits and derivatives come into play. They help reflect real-world continuity and change.
• **Simplify When Possible**: In our example, simplifying the expression for average velocity before substituting values made calculations easier. Here, consistency across any \( h \) value demonstrated that the object moved with a constant speed.

Remember, calculus provides insight into various physical and theoretical phenomena. While tackling calculus problems might seem complex, systematic approaches bring clarity and solutions.