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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 certain cable car in San Francisco can stop in \(10 \mathrm{~s}\) when traveling at maximum speed. On one occasion, the driver sees a dog a distance \(d\) in front of the car and slams on the brakes instantly. The car reaches the dog \(8.0 \mathrm{~s}\) later, and the dog jumps off the track just in time. If the car travels \(4.0 \mathrm{~m}\) beyond the position of the dog before coming to a stop, how far was the car from the dog? (Hint: You will need three equations.)

A speedboat increases its speed uniformly from \(v_{i}=\) \(20.0 \mathrm{~m} / \mathrm{s}\) to \(v_{f}=30.0 \mathrm{~m} / \mathrm{s}\) in a distance of \(2.00 \times 10^{2} \mathrm{~m}\). (a) Draw a coordinate system for this situation and label the relevant quantities, including vectors. (b) For the given information, what single equation is most appropriate for finding the acceleration? (c) Solve the equation selected in part (b) symbolically for the boat's acceleration in terms of \(v_{f}, v_{i}\), and \(\Delta x\). (d) Substitute given values, obtaining that acceleration. (e) Find the time it takes the boat to travel the given distance.

A steam catapult launches a jet aircraft from the aircraft carrier John C. Stennis, giving it a speed of \(175 \mathrm{mi} / \mathrm{h}\) in \(2.50 \mathrm{~s}\). (a) Find the average acceleration of the plane. (b) Assuming the acceleration is constant, find the distance the plane moves.

In a test run, a certain car accelerates uniformly from zero to \(24.0 \mathrm{~m} / \mathrm{s}\) in \(2.95 \mathrm{~s}\). (a) What is the magnitude of the car's acceleration? (b) How long does it take the car to change its speed from \(10.0 \mathrm{~m} / \mathrm{s}\) to \(20.0 \mathrm{~m} / \mathrm{s}\) ? (c) Will doubling the time always double the change in speed? Why?

A car starts from rest and travels for \(t_{1}\) seconds with a uniform acceleration \(a_{1} .\) The driver then applies the brakes, causing a uniform acceleration \(a_{2}\). If the brakes are applied for \(t_{2}\) seconds, (a) how fast is the cas going just before the beginning of the braking period: (b) How far does the car go before the driver begins to brake? (c) Using the answers to parts (a) and (b) as the initial velocity and position for the motion of the ca during braking, what total distance does the car travel Answers are in terms of the variables \(a_{1}, a_{2}, t_{1}\), and \(t_{2}\).
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