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Chapter 2: Problem 58

A brick is dropped from the roof of a tall building. After it has been falling for a few seconds, it falls 40.0 m in a 1.00-s time interval. What distance will it fall during the next 1.00 s? Ignore air resistance.

Short Answer

Expert verified
The brick falls 49.8 meters during the next 1.00 s.

Step by step solution

01

Understand the problem

A brick falls 40.0 meters in a second, and we need to determine how far it will fall in the next second. Assumption: no air resistance.
02

Define the key equations

We use the basic equations of motion under constant gravity, where initial velocity is denoted as \(v_i\), final velocity as \(v_f\), acceleration \(a\) (which is gravity, 9.8 m/s虏), and \(s\) is the distance.
03

Calculate the initial velocity at the start of the first 1.00 s

We use the equation \(s = v_i t + \frac{1}{2} a t^2\). Set \(s = 40.0\, m\), \(a = 9.8\, m/s^2\), and \(t = 1.0\, s\). Substitute these into the equation: \(40.0 = v_i (1) + \frac{1}{2} \times 9.8 \times 1^2\). Simplifying gives \(40.0 = v_i + 4.9\). Solve for \(v_i\): \(v_i = 35.1\, m/s\).
04

Calculate the final velocity at the end of the first 1.00 s

Use the equation \(v_f = v_i + at\). Substitute \(v_i = 35.1\, m/s\), \(a = 9.8\, m/s^2\), and \(t = 1.0\, s\): \(v_f = 35.1 + 9.8 = 44.9\, m/s\).
05

Calculate the distance fallen in the next 1.00 s

Using the equation for distance \(s = v_i t + \frac{1}{2} a t^2\), where \(t = 1.0\, s\), \(a = 9.8\, m/s^2\), and \(v_i = 44.9\, m/s\) for the next second: \(s = 44.9(1) + \frac{1}{2} \times 9.8 \times 1^2 = 44.9 + 4.9\). Therefore, \(s = 49.8\, m\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Fall
When an object is subject only to the force of gravity, it is said to be in free fall. This means that no other forces, like air resistance, are acting on it. In free fall, all objects in the same gravitational field will accelerate at the same rate, regardless of their masses.
This acceleration due to gravity is constant and can be measured depending on the location on Earth. One of the intriguing aspects of free fall is that objects accelerate at a rate of approximately 9.8 m/s虏, which we refer to as gravitational acceleration.
  • Neglects air resistance
  • Assumes constant gravitational acceleration
  • Objects of different masses fall at the same rate
Understanding these concepts helps to explore motion in a straightforward context, providing a foundational base for tackling more complex physics problems.
Equations of Motion
Equations of motion are fundamental tools in physics to describe the movement of objects. They allow us to predict the future position and velocity of an object when initial conditions are known.
These equations apply to uniformly accelerated motion, like an object in free fall, where the acceleration is constant. The key equations of motion include:
  • \[ s = v_i t + \frac{1}{2} a t^2 \]Explains distance covered
  • \[ v_f = v_i + at \]Gives the final velocity
  • \[ v_f^2 = v_i^2 + 2as \]Relates velocity and distance
In these equations,
  • \( s \) represents the distance traveled
  • \( v_i \) is the initial velocity
  • \( v_f \) is the final velocity
  • \( a \) is the acceleration
  • \( t \) is the time
Mastering these equations allows us to solve various motion-related problems with confidence.
Gravitational Acceleration
Gravitational acceleration is the acceleration experienced by an object when gravity is the only force acting on it. On Earth, this is approximately 9.8 m/s虏. It is symbolized by \( g \) and is a crucial component in calculating the motion of freely falling objects.
The magnitude of \( g \) can vary slightly depending on your location on Earth鈥攂eing slightly less at the equator than at the poles due to the planet's shape and rotation.
  • Universal for all objects in free fall
  • Standard value is 9.8 m/s虏 on Earth
  • Integral to predicting motion using equations of motion
Gravitational acceleration provides a predictable, consistent framework to solve problems of motion in a gravitational field efficiently. This understanding is paramount in fields ranging from basic physics to engineering and space exploration.

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Most popular questions from this chapter

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 m/s (45 mi/h) when it reaches the end of the 120-m-long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

A ball is thrown straight up from the ground with speed \(v_0\). At the same instant, a second ball is dropped from rest from a height \(H\), directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_0\) and \(g\) such that at the instant when the balls collide, the first ball is at the highest point of its motion.

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (\(\textbf{Fig. E2.44}\)). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_y-t\), \(v_y-t\), and \(y-t\) graphs for the motion.

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.80 m to the bottom of the incline is 3.80 m/s. What is the speed of the block when it is 3.40 m from the top of the incline?

The position of the front bumper of a test car under microprocessor control is given by \(x(t) =\) 2.17 m \(+\) (4.80 m/s\(^2)t^2\) \(-\) (0.100 m/s\(^6)t^6\). (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x-t, v_x-t\), and \(a_x-t\) graphs for the motion of the bumper between \(t =\) 0 and \(t =\) 2.00 s.
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