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Chapter 2: Problem 64

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 \(\mathrm{s}\) . 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 1796.48 m.

Step by step solution

01

Find Distance During Acceleration

The train accelerates from rest, which means its initial velocity \(v_0 = 0\). Using the formula for distance covered under constant acceleration \(s = v_0t + \frac{1}{2}at^2\), we substitute \(a = 1.60 \, \text{m/s}^2\) and \(t = 14.0 \, \text{s}\):\[ s_1 = 0 + \frac{1}{2} \times 1.60 \times (14.0)^2 \]Calculating this gives:\[ s_1 = 0 + 0.80 \times 196 \]\[ s_1 = 156.8 \, \text{m}\].
02

Determine Speed at End of Acceleration

To find the speed of the train at the end of the acceleration period, we use the formula \(v = v_0 + at\). With \(v_0 = 0\), \(a = 1.60 \, \text{m/s}^2\) and \(t = 14.0 \, \text{s}\):\[ v = 0 + 1.60 \times 14.0 \]\[ v = 22.4 \, \text{m/s}\].
03

Calculate Distance at Constant Speed

The train travels at a constant speed of 22.4 \(\text{m/s}\) for 70.0 \(\text{s}\). The distance covered at constant speed is given by the formula \(s = vt\):\[ s_2 = 22.4 \times 70.0 \]\[ s_2 = 1568.0 \, \text{m}\].
04

Find Distance During Deceleration

To find the distance covered during deceleration, we first need to know the time taken to stop. Using the formula \(v = v_0 - at\), where \(v = 0\), \(v_0 = 22.4 \, \text{m/s}\), and \(a = 3.50 \, \text{m/s}^2\), solve for \(t\):\[ 0 = 22.4 - 3.50t \]\[ t = \frac{22.4}{3.50} \approx 6.4 \, \text{s} \].Now use the distance formula \(s = v_0t - \frac{1}{2}at^2\):\[ s_3 = 22.4 \times 6.4 - \frac{1}{2} \times 3.50 \times (6.4)^2 \]\[ s_3 = 143.36 - 71.68 \]\[ s_3 = 71.68 \, \text{m}\].
05

Calculate Total Distance

Add up all the distances: \(s_1\), \(s_2\), and \(s_3\):\[ s_{\text{total}} = 156.8 + 1568.0 + 71.68 \]\[ s_{\text{total}} = 1796.48 \, \text{m}\].

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Key Concepts

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

Acceleration
When an object speeds up from rest, it experiences acceleration. In this exercise, we are given that the subway train starts from rest. This means its initial velocity, denoted as \( v_0 \), is 0 meters per second. Acceleration occurs at a rate of 1.60 m/s虏 for a duration of 14 seconds. This indicates that over each second, the train's speed increases by 1.60 meters per second.

To calculate the distance covered during acceleration, we use the kinematic formula:
  • \( s = v_0 t + \frac{1}{2} a t^2 \)
Here, \( t \) is time, and \( a \) is acceleration. Substituting the given values, the train covers a distance of 156.8 meters during acceleration. This part of motion demonstrates how forces cause change in motion, crucial in physics and real-world scenarios.
Constant Speed Motion
After accelerating, the subway train reaches a constant speed. This part of the journey involves moving at a steady pace without speeding up or slowing down. In our exercise, the train moves at a constant speed of 22.4 m/s for 70 seconds. This scenario highlights the concept of constant speed motion, where speed doesn't fluctuate.

To find the distance traveled at constant speed, we use the straightforward formula:
  • \( s = v t \)
where \( v \) is the constant velocity. By plugging in the values, the train travels 1568 meters in the constant speed section. Understanding constant speed motion helps us predict how long a journey will take when speed remains unchanged.
Deceleration
Deceleration occurs when an object slows down, or its speed decreases over time. For the subway train, deceleration happens as it approaches the next station. The train starts decelerating from a speed of 22.4 m/s and continues until it stops, with a deceleration rate of 3.50 m/s虏.

To determine how long it takes to stop, we use the equation:
  • \( 0 = v_0 - at \)
Solving for time gives us approximately 6.4 seconds. Knowing this, we can calculate the distance traveled during deceleration using the formula:
  • \( s = v_0 t - \frac{1}{2} at^2 \)
The train covers 71.68 meters as it decelerates to a complete stop. This deceleration concept is key in designing safe transport systems, ensuring sufficient stopping time and space.
Distance Calculation
Calculating total distance involves summing up the distances from each segment of motion: acceleration, constant speed, and deceleration. Each segment contributes to the entire journey.

Here's how we combine them:
  • Acceleration Distance: 156.8 meters
  • Constant Speed Distance: 1568 meters
  • Deceleration Distance: 71.68 meters
Adding these distances together, the total distance covered by the train from start to stop is 1796.48 meters. Understanding how to calculate total distance using individual motion phases offers insight into planning and analyzing travel in physics, as it combines various forms of motion into a single journey calculation.

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Most popular questions from this chapter

A Honda Civic travels in a straight line along a road. Its distance \(x\) from a stop sign is given as a function of time \(t\) by the equation \(x(t)=\alpha t^{2}-\beta t^{3},\) where \(\alpha=1.50 \mathrm{m} / \mathrm{s}^{2}\) and \(\beta=\) 0.0500 \(\mathrm{m} / \mathrm{s}^{3} .\) Calculate the average velocity of the car for each time interval: \((\mathrm{a}) t=0\) to \(t=2.00 \mathrm{s} ;\) (b) \(t=0\) to \(t=4.00 \mathrm{s}\) ; (c) \(t=2.00\) s to \(t=4.00 \mathrm{s}.\)

A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 \(\mathrm{m}\) above the ground. After an additional \(4.75 \mathrm{s},\) it is 1.00 \(\mathrm{km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 -s part of its flight and (b) the first 5.90 s of its flight.

In the vertical jump, an athlete starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let \(y_{\text { max }}\) be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above \(y_{\max } / 2\) to the time it takes him to go from the floor to that height. You may ignore air resistance.

Launch of the Space Shuttle. At launch the space shuttle weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 \(\mathrm{km} / \mathrm{h}\) , and at the end of the first 1.00 \(\mathrm{min}\) its speed is 1610 \(\mathrm{km} / \mathrm{h}\) . (a) What is the average acceleration (\(\operatorname{in}\) \(\mathrm{m} / \mathrm{s}^{2}\) ) of the shuttle (i) during the first \(8.00 \mathrm{s},\) and (ii) between 8.00 \(\mathrm{s}\) and the end of the first 1.00 \(\mathrm{min}\) ? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals, what distance does the shuttle travel (i) during the first \(8.00 \mathrm{s},\) and ( ii) during the interval from 8.00 s to 1.00 \(\mathrm{min}\)?

The acceleration of a particle is given by \(a_{x}(t)=\) \(-2.00 \mathrm{m} / \mathrm{s}^{2}+\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t .\) (a) Find the initial velocity \(v_{0 x}\) such that the particle will have the same \(x\) -coordinate at \(t=4.00 \mathrm{s}\) as it had at \(t=0 .\) (b) What will be the velocity at \(t=4.00\) s?
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