/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Collision. The engineer of a pas... [FREE SOLUTION] | 91Ó°ÊÓ

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Collision. 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 onthe same track (Fig. P2.70). 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
(a) Yes, there is a collision. (b) It occurs at 752 meters.

Step by step solution

01

Define known variables

The passenger train has an initial speed \( v_{p0} = 25.0 \ \text{m/s} \), initial position \( x_{p0} = 0 \ \text{m} \), and acceleration \( a_p = -0.100 \ \text{m/s}^2 \). The freight train has an initial speed \( v_{f} = 15.0 \ \text{m/s} \), initial position \( x_{f0} = 200 \ \text{m} \), and no acceleration (\( a_{f} = 0 \ \text{m/s}^2 \)).
02

Determine equations of motion

For the passenger train, the position as a function of time is \( x_p(t) = x_{p0} + v_{p0}t + \frac{1}{2}a_pt^2 \). For the freight train, since it moves with constant speed, the position is \( x_f(t) = x_{f0} + v_{f}t \).
03

Set equations equal to find collision time

Equate the positions of both trains to find the time \( t \) when they collide: \( x_p(t) = x_f(t) \). Plugging in the expressions gives \[ 0 + 25.0t - \frac{1}{2}(0.100)t^2 = 200 + 15.0t \].
04

Solve quadratic equation for time

Rearrange and simplify the equation to form a quadratic equation: \[ -0.05t^2 + 10t - 200 = 0 \]. Use the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -0.05 \), \( b = 10 \), and \( c = -200 \).
05

Calculate collision time values

Calculate the discriminant: \( b^2 - 4ac = 10^2 - 4(-0.05)(-200) = 100 - 40 = 60 \), which is positive, indicating two real roots. Solve: \[ t = \frac{-10 \pm \sqrt{60}}{-0.1} \], which gives two possible times.
06

Select feasible collision time

The realistic time of collision must be positive. From \( t = \frac{-10 \pm 7.75}{-0.1} \), this yields \( t_1 = 36.8 \ \text{s} \) (positive) and \( t_2 = 5.2 \ \text{s} \) (negative), select \( t = 36.8 \ \text{s} \).
07

Find collision position

Substitute \( t = 36.8 \ \text{s} \) back into either position equation. Using the freight train equation: \( x_f(t) = 200 + 15.0(t) = 200 + 15(36.8) = 752 \ \text{m} \).
08

Conclusion for part (a) and (b)

(a) Yes, a collision will occur. (b) The collision will take place at \( x = 752 \ \text{m} \).
09

Sketch position-time graph

Create a graph with time \( t \) on the horizontal axis and position \( x \) on the vertical axis. Plot the curve of \( x_p(t) = 25.0t - 0.05t^2 \) for the passenger train and the line \( x_f(t) = 200 + 15t \) for the freight train, showing intersection at \( x = 752 \ \text{m}, t = 36.8 \ \text{s} \).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that describes the motion of objects without considering the causes of this motion, such as forces. It deals with various parameters like displacement, velocity, and acceleration.
In this collision problem, the passenger train and the freight train each have distinct kinematics. The passenger train has a certain initial velocity, an acceleration applied against its direction, and it moves according to its position-time equation
\[ x_p(t) = x_{p0} + v_{p0}t + \frac{1}{2}a_pt^2 \].
The freight train, on the other hand, moves with a constant speed, so its position-time relation is linear:
\[ x_f(t) = x_{f0} + v_{f}t \].
Understanding these equations is crucial, as they allow us to plot each train’s journey over time, helping us predict where and when they might collide.
These two equations are mathematical models of their respective kinematics, representing how their positions change over time.
Quadratic Equation Solution
When analyzing the collision of two trains moving with different equations of motion, we end up needing to solve a quadratic equation. This kind of equation appears in scenarios where two objects keep changing their relative distance over time, such as with accelerating and constant-speed motion.
The quadratic equation resulting from this problem has the form:
\[ -0.05t^2 + 10t - 200 = 0 \]
which can be solved using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
Here, \( a = -0.05 \), \( b = 10 \), and \( c = -200 \).
Upon solving, we find two potential collision times. It's important to choose the realistic one, where time can only be positive, yielding that the feasible collision time is \( t = 36.8 \ \text{s} \).
The discriminant \( b^2 - 4ac \) in this context tells us the nature of the roots. A positive discriminant indicates two real roots, aligning with the physical scenario of collision prediction.
Position-Time Graph
A position-time graph is a powerful tool in physics for visualizing the motion of objects. In this collision scenario, such a graph illustrates how the position of each train changes over time.
To sketch this graph, the horizontal axis represents time \( t \) and the vertical axis stands for position \( x \).
The passenger train's path is a curve due to its acceleration opposite to its motion. Its graph follows the equation:
\[ x_p(t) = 25.0t - 0.05t^2 \],
showcasing a parabola that decreases over time as the train slows down.
The freight train moves at a constant speed, its graph is a straight line:
\[ x_f(t) = 200 + 15t \],
with a slope representing its constant velocity.
The collision point is where these two paths intersect. On the graph, it occurs at the position \( x = 752 \ \text{m} \) when \( t = 36.8 \ \text{s} \).
Drawing such graphs allows us to visually analyze and understand the motion dynamics involved in kinematics problems like collisions.

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Most popular questions from this chapter

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\). (a) How tall is the cliff? (b) If air resistance is neglected, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)

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