Question

The following table gives the position \(s(t)\) of an object moving along a line at time \(t .\) Determine the average velocities over the time intervals \([2,2.01],[2,2,001],\) and \([2,2.0001] .\) Then make a conjecture about the value of the instantancous velocity at \(t=2\) $$\begin{array}{|l|l|l|l|l|} \hline t & 2 & 2.0001 & 2.001 & 2.01 \\ \hline s(t) & 56 & 55.99959984 & 55.995984 & 55.9584 \\ \hline \end{array}$$

Short answer

Answer: The conjectured instantaneous velocity at \(t=2\) is -4.

Step-by-step solution

Find the average velocities over the given time intervals

We will use the average velocity formula to calculate the average velocities for the given time intervals: \(s(t_1)=56\) when \(t=2\) 1. For \([2, 2.01]\): \(s(t_2)=55.9584\) when \(t=2.01\) Average velocity \(=\frac{s(t_2)-s(t_1)}{t_2-t_1} = \frac{55.9584-56}{2.01-2}=-4.16\) 2. For \([2, 2.001]\): \(s(t_2)=55.995984\) when \(t=2.001\) Average velocity \(=\frac{s(t_2)-s(t_1)}{t_2-t_1} = \frac{55.995984-56}{2.001-2}=-4\) 3. For \([2, 2.0001]\): \(s(t_2)=55.99959984\) when \(t=2.0001\) Average velocity \(=\frac{s(t_2)-s(t_1)}{t_2-t_1} = \frac{55.99959984-56}{2.0001-2}=-4\)

Make a conjecture about the instantaneous velocity at \(t=2\)

As we can see, the average velocities for the intervals \([2, 2.001]\) and \([2, 2.0001]\) are both equal to -4, while for \([2, 2.01]\), the average velocity is -4.16 (which is still very close to -4). As the intervals approach a negligible difference from \(t=2\), the average velocities approach a value of -4. From this observation, we can make a conjecture that the instantaneous velocity at \(t=2\) is -4.

Key concepts

Average Velocity

When we look at the movement of an object over a period of time, we often want to know how fast it was moving on average. The average velocity helps us figure this out.
It’s the change in position \(s(t_2) - s(t_1)\) over the change in time \((t_2 - t_1)\). In simple terms, it's like saying "How much did we move, and how long did it take?"
For the problem given, average velocities are calculated for different small time intervals starting from \(t = 2\) where:

• For the interval \([2, 2.01]\), the average velocity is -4.16
• For the interval \([2, 2.001]\), it's -4
• And for \([2, 2.0001]\), it stays at -4 too

You see, as the time intervals get smaller, the average velocity becomes more precise and it appears to stabilize around -4. This provides an important clue for understanding instantaneous velocity.

Position Function

The position function, often written as \(s(t)\), describes where an object is at any given time. It’s like a map showing us the exact location or how far along a line an object has moved.
Imagine having a series of movie frames showing a car’s position every second. The position function \(s(t)\) provides the data point for each of these frames. So if you know the time, you can pinpoint the car's location.
In our exercise, the position function values at different time points help us calculate average velocities. These values are given as:

• \(s(2) = 56\)
• \(s(2.0001) = 55.99959984\)
• \(s(2.001) = 55.995984\)
• \(s(2.01) = 55.9584\)

Last, using these position values and finding the difference between them over small intervals helps us study movement more closely.

Limits

Limits are crucial in understanding instantaneous velocity. Imagine trying to capture precisely how fast an object is moving at an exact moment in time. That's what limits help us with.
They allow us to look at what happens as time intervals get incredibly small.
When we calculate average velocity over intervals like \([2, 2.0001]\), \([2, 2.001]\), or even \([2, 2.01]\) and see how this velocity "settles" to a particular value as the intervals decrease, limits are quietly doing their magic.In this exercise, the limits teach us that as the interval between the time points approaches zero, the average velocity approaches -4. Through this lens, limits guide us to make a conjecture about the instantaneous velocity.

Calculus

Calculus is the branch of mathematics that lets us study how things change. It’s essential when exploring motion and velocity. Calculus introduces us to derivatives, which are like snapshots of how fast something is changing at any given instant—hitting right at instantaneous velocity.
So, why is calculus so important here? It lets us move from average velocities over intervals to knowing exact velocity at a specific time. It’s like switching from using a flashlight to scan around a room to installing a spotlight that illuminates one small area brightly.
In our exercise, the concept of approaching the instantaneous velocity through shrinking intervals is powered by calculus. When average velocities converge as intervals shrink, calculus tells us what the exact instantaneous change is—it’s the link we need to know that our conjecture of the velocity being -4 at \((t=2)\) is correct.