Problem 66 - Precalculus Mathematics for Calculus

Question

When an object is allowed to fall freely near the surface of the earth, the gravitational pull is such that the object falls \(16 \mathrm{ft}\) in the first second, \(48 \mathrm{ft}\) in the next second, \(80 \mathrm{ft}\) in the next second, and so on. (a) Find the total distance a ball falls in 6 s. (b) Find a formula for the total distance a ball falls in \(n\) seconds.

Step-by-Step Solution

Answer

(a) 576 ft; (b) Formula: \(S_n = 16n^2\).

Step 1: Observe the pattern

Notice the sequence of distances: 16 ft, 48 ft, 80 ft,..., which increases by 32 ft each second. This indicates an arithmetic sequence with the first term (\(a_1\)) being 16 ft and a common difference (\(d\)) of 32 ft.

Step 2: Find the distance fallen in each second

We calculate the distance fallen for each second using the arithmetic sequence formula. The distance fallen during the \(n\)-th second is given by: \[ a_n = a_1 + (n-1) \cdot d = 16 + (n-1) \cdot 32 \]

Step 3: Calculate the total distance fallen in 6 seconds

Add the distances for each of the six seconds using the formula: 1st second: \(16\) ft, 2nd second: \(48\) ft, 3rd second: \(80\) ft, 4th second: \(112\) ft, 5th second: \(144\) ft, 6th second: \(176\) ft. Sum these values: \[ 16 + 48 + 80 + 112 + 144 + 176 = 576 \text{ ft} \]

Step 4: Derive the formula for total distance fallen in \(n\) seconds

The total distance for \(n\) seconds is the sum of an arithmetic sequence, which is calculated using the formula for the sum of the first \(n\) terms of an arithmetic sequence: \[ S_n = \frac{n}{2} \cdot (a_1 + a_n) \] Where \( a_n = 16 + (n-1) \cdot 32 \). Thus, \[ S_n = \frac{n}{2} \left(16 + [16 + (n-1) \cdot 32] \right) = \frac{n}{2} \left(32 + 32(n-1) \right) \] After simplification, this becomes: \[ S_n = \frac{n}{2} \cdot 32n = 16n^2 \]

Step 5: Conclude with the solutions

(a) The ball falls a total distance of 576 ft in 6 seconds. (b) The formula for the total distance fallen in \(n\) seconds is \( S_n = 16n^2 \).

Key Concepts

Gravitational Motion

Gravitational motion is the force that pulls objects towards the center of the earth. This force causes free-falling objects to accelerate at a constant rate, typically measured as 32 feet per second squared near Earth's surface. This means that every second, the object's speed increases by this amount. Gravitational motion gives us the pattern observed when objects fall, contributing to how we predict the distance they travel. In our exercise example, an object falling from rest is subject to this force, and we see an arithmetic sequence of distances fallen each second due to the uniform acceleration.

Formula Derivation

Deriving a formula for solving mathematical problems involves finding a general algebraic expression based on patterns and rules. For instance, in our exercise, the distances constitute an arithmetic sequence, identifiable through its common difference. With the first term being 16 ft and a difference of 32 ft, the formula for the distance fallen during any second is derived using\(a_n = a_1 + (n-1) \cdot d\). For total distance in \(n\) seconds, we use the sum of an arithmetic sequence formula, \(S_n = \frac{n}{2} \cdot (a_1 + a_n)\). Simplifying this expression gives \(S_n = 16n^2\), predicting the total distance fallen over \(n\) seconds.

Arithmetic Series

An arithmetic series is the sum of the terms in an arithmetic sequence. In our context, it refers to the total distance an object falls over a series of seconds. Each distance per second forms a term in the series, and with a pattern emerging, the sum of these can be calculated. The formula used is \(S_n = \frac{n}{2} \cdot (a_1 + a_n)\), which greatly simplifies calculating sums of increasing sequences. This is especially helpful in physics and mathematics, offering a clear method to handle sequences influenced by constant changes, like the uniformly accelerating object due to gravity.