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Chapter 2: Problem 62

The engineer of a passenger train traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) sights a freight train whose caboose is \(200 \mathrm{~m}\) ahead on the same track (Fig. \(\mathrm{P} 2.62\) ). The freight train is traveling at \(15.0 \mathrm{~m} / \mathrm{s}\) in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of \(0.100 \mathrm{~m} / \mathrm{s}^{2}\) in a direction opposite to the train's velocity, while the freight train continues with constant speed. Take \(x=0\) at the location of the front of the passenger train when the engineer applies the brakes. (a) Will the cows nearby witness a collision? (b) If so, where will it take place? (c) On a single graph, sketch the positions of the front of the passenger train and the back of the freight train.

Short Answer

Expert verified
No, the cows will not witness a collision. The freight train covers a distance of 3750 m in 250 seconds, which is more than the 3325 m distance that the passenger train covers before coming to a halt. Thus, the passenger train will not catch up with the freight train.

Step by step solution

01

Determine the relative speed

The relative speed of the passenger train with respect to the freight train is \(25.0 \, m/s - 15.0 \, m/s = 10.0 \, m/s\).
02

Calculate the time taken for stopping the passenger train

Using the equation of motion \(v = u - a*t\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration and \(t\) is the time, we have \(0 = 25.0 \, m/s - 0.100 \, m/s^2 * t\). Solving for \(t\), we get \(t = 25.0 \, m/s 梅 0.100 \, m/s^2 = 250 \, sec\). This is the time taken for the passenger train to come to a stop after applying brakes.
03

Calculate the distance covered by the passenger train during this time

Using the equation of motion \[s = u*t - 0.5*a*t^2\], where \(s\) is the distance, we calculate the distance covered by the passenger train during the 250 seconds after brakes are applied. Substituting \(u = 25.0 \, m/s\), \(a = 0.100 \, m/s^2)\) and \(t = 250 \, sec\), we get \(s = 25.0 \, m/s * 250 \, sec - 0.5 * 0.100 \, m/s^2 * (250 \, sec)^2 = 3125 m\). This is the distance covered by the passenger train during the time it comes to rest.
04

Calculate the distance covered by the freight train during this time

The freight train is moving at a constant speed. So, we can use the equation \(s = vt\), where \(v = 15 \, m/s\) and \(t = 250 \, sec\). Hence, \(s = 15 \, m/s * 250 \, sec = 3750 \, m\). This is the distance covered by the freight train.
05

Determine if a collision will occur

The collision will occur if the distance covered by the freight train is less than the sum of the initial distance between the two trains and the distance covered by the passenger train. The initial distance was 200 m, so the total distance for not having a collision is \(3125 \, m + 200 \, m = 3325 \, m\). Since the distance covered by the freight train is 3750 m, which is more than 3325 m, no collision will occur.
06

Graphical representation

In the graph, one axis can represent time (up to 250 sec), and the other axis represents distance. The graph for the passenger train starts at 200 m (initial distance from freight train), and ends at 3325 m (distance travelled before stopping). It slopes downward due to the decelerating passenger train. The graph for the freight train starts at 0 (initial point) and ends at 3750 m (distance travelled after 250 sec). It is a straight line due to the freight train鈥檚 constant speed.

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Key Concepts

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

Understanding Relative Speeds
When dealing with moving objects, especially in physics problems involving potential collisions, understanding relative speed is crucial. Relative speed refers to the velocity of one object as observed from another moving object. To simplify, it is how fast one object is moving in relation to another.

In our train collision problem, the passenger train's speed of 25.0 m/s and the freight train's speed of 15.0 m/s are given. The relative speed is ascertained by subtracting the speed of the freight train from the speed of the passenger train, resulting in 10.0 m/s. This calculation essentially gives us the speed at which the passenger train is closing in on the freight train.

The concept of relative speed takes a crucial role when determining whether two objects on a collision course will actually meet. In the context of our problem, by knowing the relative speed, we estimated how quickly the gap between the two trains would close, if at all, after the brakes were applied.
Equations of Motion
The equations of motion are mathematical tools that describe the relationship between an object's displacement, initial velocity, final velocity, acceleration, and time. In our example problem, these equations help us to calculate how much time it takes for the passenger train to stop and the distance it covers during this period.

We have used the equation \(v = u - a\cdot t\) to find out the time it takes for the passenger train to reach a full stop after braking. With \(v\) as the final velocity (which is 0 because the train stops), \(u\) as the initial velocity, \(a\) as the acceleration, and \(t\) as time, we rearranged this equation to solve for the time variable. The result indicates the duration in which the train will cease moving.

Another equation, \(s = u\cdot t - 0.5\cdot a\cdot t^2\), gives us the distance traveled while braking. This part is critical, as it tells us how far the passenger train moves before it stops and whether this distance is sufficient to prevent a collision with the freight train.
The Role of Constant Acceleration
Constant acceleration is when an object's velocity changes at a steady rate over time. In our train problem, once the brakes are applied, the passenger train slows down at a constant acceleration. This is a key component in solving the problem, as constant acceleration allows us to use the aforementioned equations of motion with straightforward arithmetic rather than having to involve calculus which is necessary for non-constant acceleration.

Here, the negative sign of the acceleration value, \(-0.100\,\mathrm{m/s}^2\), signifies that the train is slowing down. The beauty of constant acceleration is the predictability it allows in calculating motion. We can determine exactly when and where an object will be at any moment, which we did to confirm that the passenger train will not collide with the freight train over the given distance.

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Most popular questions from this chapter

A lunar lander is making its descent to Moon Base I (Fig. E2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is \(5.0 \mathrm{~m}\) above the surface and has a downward speed of \(0.8 \mathrm{~m} / \mathrm{s}\). With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is \(1.6 \mathrm{~m} / \mathrm{s}^{2}\).

You are standing at rest at a bus stop. A bus moving at a constant speed of \(5.00 \mathrm{~m} / \mathrm{s}\) passes you. When the rear of the bus is \(12.0 \mathrm{~m}\) past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of \(0.960 \mathrm{~m} / \mathrm{s}^{2}\). How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, \(6.00 \mathrm{~s}\) after it was thrown. What is the speed of the rock just before it reaches the water \(28.0 \mathrm{~m}\) below the point where the rock left your hand? Ignore air resistance.

A large boulder is ejected vertically upward from a volcano with an initial speed of \(40.0 \mathrm{~m} / \mathrm{s}\). Ignore air resistance. (a) At what time after being ejected is the boulder moving at \(20.0 \mathrm{~m} / \mathrm{s}\) upward? (b) At what time is it moving at \(20.0 \mathrm{~m} / \mathrm{s}\) downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion.

An object's velocity is measured to be \(v_{x}(t)=\alpha-\beta t^{2}\), where \(\alpha=4.00 \mathrm{~m} / \mathrm{s}\) and \(\beta=2.00 \mathrm{~m} / \mathrm{s}^{3} .\) At \(t=0\) the object is at \(x=0 .\) (a) Calculate the object's position and acceleration as functions of time. (b) What is the object's maximum positive displacement from the origin?
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