Problem 3 - Applied Calculus

Question

In a time of \(t\) seconds, a particle moves a distance of \(s\) meters from its starting point, where \(s=3 t^{2}\). (a) Find the average velocity between \(t=1\) and \(t=\) \(1+h\) if: (i) \(h=0.1\), (ii) \(h=0.01\), (iii) \(h=\underline{0.001 \text { . }}\) (b) Use your answers to part (a) to estimate the instantaneous velocity of the particle at time \(t=1\).

Step-by-Step Solution

Answer

The instantaneous velocity at \(t=1\) is approximately \(6\) m/s.

Step 1: Understanding the formula for average velocity

The average velocity of a particle over a time interval from \(t=a\) to \(t=b\) is given by the formula: \( v_{avg} = \frac{s(b) - s(a)}{b - a} \), where \(s(t) = 3t^2\). For part (a), we need to calculate this average velocity for different small intervals around \(t=1\).

Step 2: Calculating distances for different times

Let's calculate \(s(1)\) and \(s(1+h)\) for the given \(h\) values. - \(s(1) = 3(1)^2 = 3\). - For \(h = 0.1\), \(s(1+0.1) = 3(1.1)^2 = 3 \times 1.21 = 3.63\).- For \(h = 0.01\), \(s(1+0.01) = 3(1.01)^2 = 3 \times 1.0201 = 3.0603\).- For \(h = 0.001\), \(s(1+0.001) = 3(1.001)^2 = 3 \times 1.002001 = 3.006003\).

Step 3: Calculating average velocity for h = 0.1

Using the average velocity formula, for \(h=0.1\):\[v_{avg} = \frac{s(1+0.1) - s(1)}{0.1} = \frac{3.63 - 3}{0.1} = \frac{0.63}{0.1} = 6.3\text{ m/s}.\]

Step 4: Calculating average velocity for h = 0.01

Similarly, for \(h=0.01\):\[v_{avg} = \frac{s(1+0.01) - s(1)}{0.01} = \frac{3.0603 - 3}{0.01} = \frac{0.0603}{0.01} = 6.03\text{ m/s}.\]

Step 5: Calculating average velocity for h = 0.001

Finally, for \(h=0.001\):\[v_{avg} = \frac{s(1+0.001) - s(1)}{0.001} = \frac{3.006003 - 3}{0.001} = \frac{0.006003}{0.001} = 6.003\text{ m/s}.\]

Step 6: Estimating instantaneous velocity at t = 1

As \(h\) approaches zero, the average velocity approaches the instantaneous velocity. From the calculations: For \(h = 0.1, 0.01, 0.001\), the average velocities are \(6.3\), \(6.03\), and \(6.003\) respectively. So, by estimating, the instantaneous velocity at \(t=1\) is approximately \(6\text{ m/s}\).

Key Concepts

Average Velocity

The concept of average velocity helps us understand the overall rate of change in a particle's position over a specified time interval. It’s calculated by dividing the total change in position (distance covered) by the total time taken. In mathematical terms, we express this as:

• Average velocity, \( v_{avg} \), is given by:\[v_{avg} = \frac{s(b) - s(a)}{b - a}\]where \(s(t)\) represents the distance function, and \(a\) and \(b\) represent the start and end points of the time interval respectively.
• For our specific exercise, the particle's distance \(s\) with respect to time \(t\) is \(s = 3t^2\).

This formula implies that to find the average velocity of the particle between two points in time, you find the difference in distance at these two times and divide it by the time difference. For very small intervals, this average can give us insights into the instantaneous velocity, especially as the time interval becomes negligible. Understanding average velocity is crucial to comprehend how motion calculations transform as you begin approximating instantaneous rates.

Particle Motion

Particle motion refers to the movement of a particle (or an object) and is quantified via its position and velocity as it changes over time. In physics, examining how a particle moves can reveal a lot about the forces and directions it experiences.

• In our exercise, the particle moves according to the distance function \(s = 3t^2\). This quadratic relationship is key to understanding the motion of the particle.
• Since \(s(t)\) is quadratic, it means that as time increases, the distance covered by the particle doesn’t just increase linearly but at an accelerating rate. The square of the time value increases the impact on the distance.

Therefore, when looking at particle motion, it’s vital to consider the nature of the function depicting its displacement relative to time. This form of analysis allows us to better understand not just the path followed but also how quickly or slowly that path is traversed, which leads us directly into grasping instantaneous velocity as the time interval diminishes.

Distance-Time Relationship

The distance-time relationship describes how the distance covered by a particle varies over time. It's an essential aspect of understanding motion since it provides insights into speed and velocity.

• In our exercise example, the distance a particle moves over time is given by the function \(s = 3t^2\).
• This formula tells us that the particle's movement is not constant; instead, its travel distance changes with the square of time, making the motion non-linear.

Understanding this relationship is crucial for determining how fast or slow a particle is moving at any given point. The steeper the curve (rate of change), the faster the particle is moving at that moment. Recognizing how the distance-time curve behaves helps in predicting future positions or retroactively understanding past movement details. When we're estimating instantaneous velocity, this relationship becomes particularly important because as you examine smaller and smaller intervals, this offers approaches to find the precise speed at an exact point in time.