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Chapter 2: Problem 37

A train is traveling down a straight track at \(20 \mathrm{~m} / \mathrm{s}\) when the engineer applies the brakes, resulting in an acceleration of \(-1.0 \mathrm{~m} / \mathrm{s}^{2}\) as long as the train is in motion. How far does the train move during a 40 -s time interval starting at the instant the brakes are applied?

Short Answer

Expert verified
The train does not move far during the 40-second time interval after the brakes are applied; the displacement is 0 meters.

Step by step solution

01

Identify and write down the known quantities

You know the initial velocity \(vi = 20 m/s\), the acceleration \(a = -1.0 m/s^2\) (since the train is decelerating), and the time \(t = 40 s\).
02

Apply the second kinematic equation

Using the second equation of motion for displacement with uniform acceleration, given as \(d = vit + 0.5at^2\), substitute the known quantities. Thus, \(d = (20 m/s)(40 s) + 0.5(-1.0 m/s^2)(40 s)^2\). Here, the first term represents the distance the train would have covered if it kept moving at the original velocity and the second term shows how the distance covered changes due to acceleration.
03

Solve for displacement

Your displacement \(d\) becomes \(d = 800 m - 800 m = 0 m\). Thus, the displacement of the train during the 40-second time interval is 0 meters.

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Key Concepts

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

Understanding Uniform Acceleration
Uniform acceleration occurs when an object's acceleration remains constant throughout its motion. In the context of the train problem, uniform acceleration means the train's speed changes at a steady rate due to the constant braking force applied by the engineer.

* Key points about uniform acceleration include:
  • The acceleration does not change over time.
  • It can be positive (speeding up) or negative (slowing down), depending on the direction.
In this exercise, the acceleration is negative, indicating the train is decelerating.
This constant deceleration implies that every second, the train's speed decreases by 1 m/s. Therefore, uniform acceleration is crucial to determining the train's behavior over time in the scenario presented.
Decoding Displacement
Displacement refers to the change in position of an object in a specific direction. It's important to note that displacement is different from distance; while distance is a scalar quantity representing the total path covered, displacement is a vector quantity and considers direction.

For the train exercise:
  • The train starts from a point and ends at the same point after 40 seconds.
  • This means that despite covering some distance forward and backward, the net change in position is zero.
The formula used, which incorporates initial velocity, time, and acceleration, helps calculate this net change. Although the train moves, the displacement is zero because the forward motion due to initial speed is exactly countered by the backward motion caused by the braking effect.
Exploring Equations of Motion
The equations of motion are fundamental in solving kinematic problems, especially when dealing with uniform acceleration. These equations link displacement, velocity, acceleration, and time.For this problem, the second equation of motion is particularly useful:
  • The equation: \( d = v_i t + \frac{1}{2} a t^2 \) combines initial velocity, time, and acceleration to find displacement.
  • The term \( v_i t \) calculates how far the train would move without any acceleration.
  • The term \( \frac{1}{2} a t^2 \) adjusts this distance to account for the train's acceleration over time.
By applying this formula, we can see the distinct effect of the train's deceleration over 40 seconds. Ultimately, these equations play a critical role in predicting how an object moves under uniform acceleration conditions.

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Most popular questions from this chapter

A truck tractor pulls two trailers, one behind the other, at a constant speed of \(100 \mathrm{~km} / \mathrm{h}\). It takes \(0.600 \mathrm{~s}\) for the big rig to completely pass onto a bridge \(400 \mathrm{~m}\) long. For what duration of time is all or part of the trucktrailer combination on the bridge?

A bullet is fired through a board \(10.0 \mathrm{~cm}\) thick in such a way that the bullet's line of motion is perpendicular to the face of the board. If the initial speed of the bullet is \(400 \mathrm{~m} / \mathrm{s}\) and it emerges from the other side of the board with a speed of \(300 \mathrm{~m} / \mathrm{s}\), find (a) the acceleration of the bullet as it passes through the board and (b) the total time the bullet is in contact with the board.

A model rocket is launched straight upward with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It accelerates with a constant upward acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) until its engines stop at an altitude of \(150 \mathrm{~m}\). (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

A car accelerates uniformly from rest to a speed of \(40.0 \mathrm{mi} / \mathrm{h}\) in \(12.0 \mathrm{~s}\). Find (a) the distance the car travels during this time and (b) the constant acceleration of the car.

An object moves with constant acceleration \(4.00 \mathrm{~m} / \mathrm{s}^{2}\) and over a time interval reaches a final velocity of \(12.0 \mathrm{~m} / \mathrm{s}\). (a) If its original velocity is \(6.00 \mathrm{~m} / \mathrm{s}\), what is its displacement during the time interval? (b) What is the distance it travels during this interval? (c) If its original velocity is \(-6.00 \mathrm{~m} / \mathrm{s}\), what is its displacement during this interval? (d) What is the total distance it travels during the interval in part (c)?
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