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Chapter 2: Problem 40

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 s. It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

Short Answer

Expert verified
The total distance covered is 1796.5 meters.

Step by step solution

01

Calculate the distance during acceleration

First, find the distance covered while the train accelerates. Use the formula for uniformly accelerated motion: \[ d_1 = v_0 t + \frac{1}{2} a t^2 \] where initial velocity \( v_0 = 0 \), acceleration \( a = 1.60 \, \mathrm{m/s}^2 \), and time \( t = 14.0 \, \mathrm{s} \). Thus, \[ d_1 = 0 + \frac{1}{2} \times 1.60 \, \mathrm{m/s}^2 \times (14.0 \, \mathrm{s})^2 = 156.8 \, \mathrm{m} \].
02

Calculate the final velocity after acceleration

Find the final velocity at the end of the acceleration period using: \[ v = v_0 + at \]. Since \( v_0 = 0 \), \( a = 1.60 \, \mathrm{m/s}^2 \), and \( t = 14.0 \, \mathrm{s} \), we have \[ v = 0 + 1.60 \times 14.0 = 22.4 \, \mathrm{m/s} \].
03

Calculate the distance at constant speed

With the train now running at a constant speed of \( v = 22.4 \, \mathrm{m/s} \) for \( 70.0 \, \mathrm{s} \), the distance covered is: \[ d_2 = v \times t = 22.4 \, \mathrm{m/s} \times 70.0 \, \mathrm{s} = 1568 \, \mathrm{m} \].
04

Calculate the distance during deceleration

For deceleration, use the formula: \[ v^2 = u^2 + 2as \], where \( v = 0 \), \( u = 22.4 \, \mathrm{m/s} \), and \( a = -3.50 \, \mathrm{m/s}^2 \). Solving for \( s \) (distance), we get: \[ 0 = (22.4)^2 + 2(-3.50)s \]. This simplifies to: \[ s = \frac{(22.4)^2}{2 \times 3.50} = 71.7 \, \mathrm{m} \].
05

Calculate the total distance

Add up all the distances covered during acceleration, constant speed, and deceleration: \[ \text{Total distance} = d_1 + d_2 + s = 156.8 + 1568 + 71.7 = 1796.5 \, \mathrm{m} \].

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Key Concepts

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

Uniformly Accelerated Motion
Uniformly accelerated motion is a key aspect of kinematics where an object's velocity changes at a constant rate. This implies that the same amount of speed is gained or lost each second. It simplifies many calculations since the acceleration remains constant throughout.
In the scenario provided, the subway train begins from rest, which means its initial velocity is zero. The train accelerates at a rate of 1.60 m/s虏 for 14 seconds. To calculate the distance covered during this phase, we use:
  • The formula: \[ d = v_0 t + \frac{1}{2} a t^2 \]
  • Plug in the values: initial velocity \( v_0 = 0 \), acceleration \( a = 1.60 \, \mathrm{m/s}^2 \), and time \( t = 14 \, \mathrm{s} \).
This formula helps to find that the train covers 156.8 meters during its acceleration period.
Velocity
Velocity describes the speed of an object in a specific direction. Unlike speed, velocity is a vector quantity, meaning it has both magnitude and direction.
In the context of the train, after accelerating for 14 seconds at 1.60 m/s虏, its velocity increases. We calculate the train's final velocity at the end of the acceleration using:
  • The formula: \[ v = v_0 + at \]
  • Substitute the values: initial velocity \( v_0 = 0 \), acceleration \( a = 1.60 \, \mathrm{m/s}^2 \), and time \( t = 14 \, \mathrm{s} \).
After evaluation, we find the train achieves a velocity of 22.4 m/s before it begins travelling at a constant speed.
Deceleration
Deceleration is the process of slowing down or decreasing velocity. It is also a form of acceleration but in the opposite direction of motion. During deceleration, an object loses speed until it comes to a stop.
For the subway train, deceleration occurs after travelling at constant speed, as it approaches the next station. The rate of deceleration here is -3.50 m/s虏. Although it sounds complex, calculating the distance during this phase uses a simple formula:
  • The formula: \[ v^2 = u^2 + 2as \]
  • Since final velocity \( v = 0 \), initial velocity \( u = 22.4 \, \mathrm{m/s} \), and deceleration \( a = -3.50 \, \mathrm{m/s}^2 \).
By solving this equation, we determine that the train covers an additional 71.7 meters as it decelerates to a stop.
Distance Calculation
Calculating total distance in motion includes three key components: distance during acceleration, distance at constant speed, and distance during deceleration.
Here's a breakdown of how we found the total distance for the subway train:
  • Acceleration phase: Covered 156.8 meters.
  • Constant speed phase: The train runs at 22.4 m/s for 70 seconds, which equals 1568 meters.
  • Deceleration phase: Calculated a distance of 71.7 meters until the train stops.
Adding these components together, the total distance covered is 1796.5 meters, demonstrating how kinematic equations can comprehensively describe complex motion scenarios.

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Most popular questions from this chapter

A cat drops from a shelf 4.0 ft above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12 \(\mathrm{cm} .\) Calculate (a) his speed when he first touches the floor (ignore air resistance), (b) how long it takes him to stop, and (c) his acceleration (assumed constant) while he is stopping, in \(\mathrm{m} / \mathrm{s}^{2}\) and \(g^{\prime} \mathrm{s}\) .

Earthquake waves. Earthquakes produce several types of shock waves. The best known are the P-waves (P for primary or pressure) and the S-waves (S for secondary or shear). In the earth's crust, P-waves travel at around 6.5 \(\mathrm{km} / \mathrm{s}\) while S-waves move at about 3.5 \(\mathrm{km} / \mathrm{s}\) . (The actual speeds vary with the type of material the waves are going through.) The time delay between the arrival of these two types of waves at a seismic recording station tells geologists how far away the earthquake that produced the waves occurred. (a) If the time delay at a seismic station is 33 s, how far from that station did the earthquake that produced the waves occurred. (a) If the time delay at a seismic station is 33 s, how far from that station did the earthquake occur? (b) One form of earthquake warning system detects the faster (but less damaging) P-waves and sounds an alarm when they first arrive, giving people a short time to seek cover before the more dangerous S-waves arrive. If an earthquake occurs 375 \(\mathrm{km}\) away from such a warning device, how much time would people have to take cover between the alarm and the arrival of the S-waves?

Prevention of hip fractures. Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip's speed at impact is about 2.0 \(\mathrm{m} / \mathrm{s}\) . If this can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) or less, the hip will usually not fracture. One way to do this is by wearing elastic hip pads. (a) If a typical pad is 5.0 \(\mathrm{cm}\) thick and compresses by 2.0 \(\mathrm{cm}\) during the impact of a fall, what acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g^{\prime}\) s) does the hip undergo to reduce its speed to 1.3 \(\mathrm{m} / \mathrm{s} ?\) (b) The acceleration you found in part (a) may seem like a rather large acceleration, but to fully assess its effects on the hip, calculate how long it lasts.

While driving on the freeway at \(110 \mathrm{km} / \mathrm{h},\) you pass a truck whose total length you estimate at 25 \(\mathrm{m}\) . (a) If it takes you, in the driver's seat, 5.5 \(\mathrm{s}\) to pass from the rear of the truck to its front, what is the truck's speed relative to the road? (b) How far does the truck travel while you're passing it?

\(\cdot\) If a pilot accelerates at more than 4\(g\) , he begins to "gray out," but not completely lose consciousness. (a) What is the shortest time that a jet pilot starting from rest can take to reach Mach 4 (four times the speed of sound) without graying out? (b) How far would the plane travel during this period of acceleration? (Use 331 \(\mathrm{m} / \mathrm{s}\) for the speed of sound in cold air.)
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