Problem 46 - Calculus Volume 1

Question

The distance in feet that a ball rolls down an incline is modeled by the function \(s(t)=14 t^{2},\) where \(t\) is seconds after the ball begins rolling. a. Find the average velocity of the ball over the following time intervals: i. [5,5.1] ii. [5,5.01] iii. \(\quad[5,5.001]\) iv. [5,5.0001] b. Use the answers from a. to draw a conclusion about the instantaneous velocity of the ball at \(t=5\) seconds.

Step-by-Step Solution

Answer

The instantaneous velocity of the ball at \( t=5 \) seconds is 140 ft/s.

Step 1: Calculate s(t) for each endpoint of the intervals

To find the average velocity of the ball over the given intervals, we will first calculate the function \( s(t) = 14t^2 \) for each endpoint of the intervals. For interval [5,5.1]:- \( s(5) = 14 \times 5^2 = 350 \)- \( s(5.1) = 14 \times (5.1)^2 = 14 \times 26.01 = 364.14 \)For interval [5,5.01]:- \( s(5) = 350 \)- \( s(5.01) = 14 \times (5.01)^2 = 14 \times 25.1001 = 351.4014 \)For interval [5,5.001]:- \( s(5) = 350 \)- \( s(5.001) = 14 \times (5.001)^2 = 14 \times 25.010001 = 350.140014 \)For interval [5,5.0001]:- \( s(5) = 350 \)- \( s(5.0001) = 14 \times (5.0001)^2 = 14 \times 25.00100001 = 350.0140014 \)

Step 2: Calculate the change in distance for each interval

Now we will calculate the change in distance \( \Delta s \) over each interval by subtracting the values calculated at \( t = 5 \) from those at \( t = 5 + \Delta t \).For interval [5,5.1]:- \( \Delta s = s(5.1) - s(5) = 364.14 - 350 = 14.14 \)For interval [5,5.01]:- \( \Delta s = s(5.01) - s(5) = 351.4014 - 350 = 1.4014 \)For interval [5,5.001]:- \( \Delta s = s(5.001) - s(5) = 350.140014 - 350 = 0.140014 \) For interval [5,5.0001]:- \( \Delta s = s(5.0001) - s(5) = 350.0140014 - 350 = 0.0140014 \)

Step 3: Calculate average velocity for each interval

The average velocity over each interval is calculated as the ratio of the change in distance to the change in time \( \Delta t \). For interval [5,5.1]:- \( \Delta t = 0.1 \)- Average Velocity \( \approx \frac{14.14}{0.1} = 141.4 \) ft/sFor interval [5,5.01]:- \( \Delta t = 0.01 \)- Average Velocity \( \approx \frac{1.4014}{0.01} = 140.14 \) ft/sFor interval [5,5.001]:- \( \Delta t = 0.001 \)- Average Velocity \( \approx \frac{0.140014}{0.001} = 140.014 \) ft/sFor interval [5,5.0001]:- \( \Delta t = 0.0001 \)- Average Velocity \( \approx \frac{0.0140014}{0.0001} = 140.0014 \) ft/s

Step 4: Draw conclusion about the instantaneous velocity at t=5

As we narrow the time intervals closer to \( t = 5 \), the average velocity approaches a limiting value. The instantaneous velocity at a point is equivalent to this limiting value. From calculations above, as the interval narrow to \( t = 5 \), the average velocity approaches 140 ft/s. Thus, the instantaneous velocity of the ball at \( t = 5 \) seconds is 140 ft/s.

Key Concepts

Average Velocity

Average velocity is an essential concept in understanding motion. When we calculate average velocity over a specific time interval, we are determining how fast an object is moving on average between two points in time. It is calculated as the total change in distance divided by the total change in time.
In this exercise, we calculated the average velocity for different time intervals. To do this, we:

• Calculated the distance at each endpoint of the interval using the given distance-time function.
• Found the difference between these distances to get the total distance traveled.

Average velocity tells us about the motion over a finite period, but it doesn't show how velocity may change from one moment to the next. As intervals become smaller, average velocity can hint at the behavior of instantaneous velocity.

Derivative

The derivative is a fundamental tool in calculus used to find the instantaneous rate of change of a function. In simpler terms, it helps us find how a quantity changes at a particular point. For instance, if a car's distance over time is described by a function, the derivative gives us its speed or velocity at any particular moment.
In the context of this exercise, the derivative of the distance-time function, which is given as \(s(t) = 14t^2\), can be calculated. The derivative, represented as \(s'(t)\), shows the rate of change of the distance with respect to time, giving us the velocity function.
To find the derivative of \(s(t) = 14t^2\), we apply the power rule of derivatives:

• The derivative of \(t^2\) is \(2t\).
• Thus, \(s'(t) = 14 imes 2t = 28t\).

So, the velocity of the ball at any time \(t\) is given by \(28t\). This derivative allows us to calculate instantaneous velocity at any chosen moment.

Limits

In calculus, limits help us understand the behavior of functions as they approach a particular point. With the limit, we focus on what happens as the differences in our values become infinitesimally small. In finding instantaneous velocity, the concept of limits proves crucial.
From our initial exploration of average velocities over shrinking intervals, the limit essentially works by making these intervals approach zero. As the interval between the two points of time becomes vanishingly small, the average velocity tends toward the instantaneous velocity.
In this specific exercise, limit allows us to conclude that as \( \Delta t \to 0 \), the average velocity approaches the instantaneous velocity. Analytically, we can say:

• \( \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = 140 \text{ ft/s}\).

Therefore, limits enable us to bridge the gap between average and instantaneous velocity by focusing on infinitesimally small time intervals.

Distance-Time Function

The distance-time function is a mathematical representation of how distance changes over time for a moving object. In this specific scenario, the function provided is \(s(t) = 14t^2\).
This quadratic function tells us that the distance the ball rolls is proportional to the square of the time since it began rolling. Such functions are typical for objects under constant acceleration.
Distance-time functions are foundational to understanding motion, especially in physics. They allow us to:

• Predict future positions by plugging different time values into the function.
• Analyze motion characteristics, like acceleration and velocity, using calculus tools such as derivatives.

Here, the function \(s(t) = 14t^2\) also illustrates how distance rapidly increases with time, which is characteristic of rolling or accelerating objects. It's crucial for solving problems related to motion, such as finding average and instantaneous velocity.