Problem 50 - Calculus and its applications

Question

(See Exercise 49.) Suppose a worker drops a bolt from a scaffold high above a work site. Assuming negligible air resistance, find (a) how far the bolt falls in \(2 \mathrm{sec},\) (b) how fast the bolt is traveling 2 sec after being dropped, and (c) the bolt's acceleration after it has been falling for 2 sec.

Step-by-Step Solution

Answer

(a) 19.6 m, (b) 19.6 m/s, (c) 9.8 m/s².

Step 1: Understand the problem

The problem involves a bolt falling from a height due to gravity. We assume no air resistance and are asked to find the distance fallen in 2 seconds, the speed at 2 seconds, and the acceleration at that time.

Step 2: Apply the formula for distance fallen

The formula for the distance traveled under constant acceleration due to gravity is \( s = ut + \frac{1}{2}gt^2 \), where \( u \) (initial velocity) is 0 m/s, \( g \) (acceleration due to gravity) is \( 9.8 \text{ m/s}^2 \), and \( t \) is the time in seconds.

Step 3: Calculate distance for 2 seconds

Using \( s = 0 \cdot t + \frac{1}{2} \times 9.8 \times (2)^2 \): \( s = \frac{1}{2} \times 9.8 \times 4 = 19.6 \text{ m} \). The bolt falls 19.6 meters in 2 seconds.

Step 4: Apply formula for velocity

Velocity after a certain time is given by \( v = u + gt \). Since the initial velocity \( u = 0 \), \( v = 9.8 \times 2 = 19.6 \text{ m/s} \).

Step 5: Determine acceleration after 2 seconds

The acceleration of an object falling freely under gravity is constant and equals \( 9.8 \text{ m/s}^2 \), regardless of the time elapsed.

Key Concepts

Acceleration Due to Gravity

When considering objects in free fall, the concept of "acceleration due to gravity" is crucial. On Earth, this acceleration is commonly approximated to be a constant value of approximately 9.8 meters per second squared ( \(9.8 \text{ m/s}^2 \) ). This means that, every second, an object's velocity increases by 9.8 m/s as it falls, provided there's no air resistance.
The constant acceleration implies that no matter how long an object has been falling, its acceleration due to gravity remains unchanged.

• This is because gravity is a uniform force acting equally at all points close to Earth's surface.
• It is important to note that this value can slightly vary depending on geographical location or altitude.

For objects just released or dropped from rest (with no initial velocity), the acceleration due to gravity is the sole force acting upon them, dictating how they move. This uniform acceleration shapes the equations used in calculating both distance fallen and velocity over time.

Distance Fallen

To find out how far an object falls under gravity, we use the formula for the distance traveled:\[ s = ut + \frac{1}{2}gt^2 \]

• Here, \( u \) represents the initial velocity, which is zero when an object starts from rest.
• \( g \) is the acceleration due to gravity, \(9.8 \text{ m/s}^2 \).
• \( t \) is the time elapsed.

This formula stems from the basic principles of kinematics. For a falling object with no initial velocity, the formula simplifies to \( s = \frac{1}{2}gt^2 \), indicating that the distance fallen is proportional to the square of the time elapsed. This is why after 2 seconds of free fall, the bolt falls 19.6 meters. The quadratic relation means that doubling the time doesn't just double the distance fallen; it increases it fourfold.

Velocity Calculation

Velocity of a falling object can be determined using the equation:\[ v = u + gt \]In cases where an object begins from rest, \( u = 0 \), leading to \( v = gt \). Here:

• \(v\) is the velocity of the object at time\(t\),
• and \(g\) is the gravitational pull on it, \(9.8 \text{ m/s}^2 \).

This tells us that the object's speed increases linearly over time. At 2 seconds, the bolt's velocity would be \(19.6 \text{ m/s} \), calculated by multiplying the acceleration due to gravity by time \(g \times t\).Understanding how velocity works in free fall situations helps illuminate the effect of gravity and how it linearly adds speed every second.