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Chapter 2: Problem 7

Two trains, each having a speed of \(30 \mathrm{~km} / \mathrm{h}\), are headed at each other on the same straight track. A bird that can fly \(60 \mathrm{~km} / \mathrm{h}\) flies off the front of one train when they are \(60 \mathrm{~km}\) apart and heads directly for the other train. On reaching the other train, the bird flies directly back to the first train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide?

Short Answer

Expert verified
The bird travels 60 km before the trains collide.

Step by step solution

01

Calculate Time Until Collision

To find the time until the trains collide, we add their speeds since they are moving towards each other: \(30 + 30 = 60\) km/h. Given the trains are 60 km apart, we calculate the time until collision as:\[ \text{Time until collision} = \frac{\text{Distance}}{\text{Relative speed}} = \frac{60 \text{ km}}{60 \text{ km/h}} = 1 \text{ hour}.\]
02

Calculate the Bird's Total Travel Time

Since the trains will collide in 1 hour, the bird also will fly for the same duration, which is 1 hour.
03

Calculate the Bird's Total Distance

The bird flies at 60 km/h. So, the total distance the bird travels is: \[ \text{Bird's total distance} = \text{Bird speed} \times \text{Time flying} = 60 \text{ km/h} \times 1 \text{ hour} = 60 \text{ km}. \]

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that studies how objects move. It explores variables such as velocity, acceleration, and time without considering the causes of motion (forces). In this exercise, we observe how the trains and the bird move along the same path.
In problems like these, the concept of relative speed plays a crucial role. When two objects move towards each other, their speeds add up to determine the relative speed. This is because both objects close the gap between each other more quickly. For example, two trains each moving at 30 km/h toward one another have a combined relative speed of 60 km/h.
The relative speed sets the stage for calculating time-based questions, such as determining when two objects will meet or collide. This is fundamental to solving problems involving moving objects on the same line of motion.
Motion
Motion refers to a change in position of an object over time. In our exercise, the motion involves trains and a bird. Each train moves at 30 km/h towards each other, and a bird continuously flies back and forth between them at a speed of 60 km/h.
This scenario highlights how different motions can be analyzed together. The bird's back-and-forth motion shows repetitive displacement in opposite directions, which adds up over time. Understanding motion allows us to predict future positions and behaviors of moving objects.
To assess such motion, consider not just the speed but also the direction. The direction plays a key role in understanding how objects travel and how their paths may intersect.
Distance Calculation
Distance calculation is essential to determine how far objects travel over a given period. In the bird and train problem, calculating distance involves multiplying speed by time.
For the bird, which flies at 60 km/h for a total of 1 hour (the time until the trains collide), it travels 60 km. Calculating distance becomes straightforward when you know the object's speed and the time it remains in motion.
The formula for this simple calculation is:
  • Distance = Speed \(\times\) Time
Calculate distance efficiently by understanding the relationship between speed, distance, and time. Such calculation is ubiquitous in physics and crucial for solving real-world problems.

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Most popular questions from this chapter

The brakes on your car can slow you at a rate of \(5.2 \mathrm{~m} / \mathrm{s}^{2}\). (a) If you are going \(137 \mathrm{~km} / \mathrm{h}\) and suddenly see a state trooper, what is the minimum time in which you can get your car under the 90 \(\mathrm{km} / \mathrm{h}\) speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.) (b) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for such a slowing.

Shows part of a street where traffic flow is to be controlled to allow a platoon of cars to move smoothly along the street. Suppose that the platoon leaders have just reached intersection 2, where the green appeared when they were distance \(d\) from the intersection. They continue to travel at a certain speed \(v_{p}\) (the speed limit) to reach intersection 3, where the green appears when they are distance \(d\) from it. The intersections are separated by distances \(D_{23}\) and \(D_{12}\). (a) What should be the time delay of the onset of green at intersection 3 relative to that at intersection 2 to keep the platoon moving smoothly? Suppose, instead, that the platoon had been stopped by a red light at intersection \(1 .\) When the green comes on there, the leaders require a certain time \(t_{r}\) to respond to the change and an additional time to accelerate at some rate \(a\) to the cruising speed \(v_{p} .\) (b) If the green at intersection 2 is to appear when the leaders are distance \(d\) from that intersection, how long after the light at intersection 1 turns green should the light at intersection 2 turn green?

An electron has a constant acceleration of \(+3.2 \mathrm{~m} / \mathrm{s}^{2}\). At a certain instant its velocity is \(+9.6 \mathrm{~m} / \mathrm{s}\). What is its velocity (a) \(2.5 \mathrm{~s}\) earlier and (b) \(2.5\) s later?

If the maximum acceleration that is tolerable for passengers in a subway train is \(1.34 \mathrm{~m} / \mathrm{s}^{2}\) and subway stations are located \(806 \mathrm{~m}\) apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for \(20 \mathrm{~s}\) at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph \(x, v\), and \(a\) versus \(t\) for the interval from one start-up to the next.

A certain elevator cab has a total run of \(190 \mathrm{~m}\) and a maximum speed of \(305 \mathrm{~m} / \mathrm{min}\), and it accelerates from rest and then back to rest at \(1.22 \mathrm{~m} / \mathrm{s}^{2} .\) (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop \(190 \mathrm{~m}\) run, starting and ending at rest?
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