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Chapter 2: Problem 34

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 meters.

Step by step solution

01

Calculate acceleration phase distance

During the acceleration phase, the train accelerates from rest. We can use the kinematic equation for distance under constant acceleration: \[ d_1 = \frac{1}{2}at^2 \]where \(a = 1.60\, \mathrm{m/s^2}\) and \(t = 14.0\, \mathrm{s}\). Substitute these values to find \(d_1\):\[ d_1 = \frac{1}{2} \times 1.60 \times (14.0)^2 = 156.8\, \mathrm{m} \]
02

Calculate constant speed phase distance

First, find the final velocity after acceleration using:\[ v = at \]with \( a = 1.60\, \mathrm{m/s^2} \) and \( t = 14.0\, \mathrm{s} \). So,\[ v = 1.60 \times 14.0 = 22.4 \mathrm{m/s} \]Now, compute the distance covered at constant speed:\[ d_2 = v \times t \]where \( v = 22.4 \mathrm{m/s} \) and \( t = 70.0 \mathrm{s} \). Thus,\[ d_2 = 22.4 \times 70.0 = 1568.0 \mathrm{m} \]
03

Calculate deceleration phase distance

For the deceleration phase, final velocity reduces to zero. Use the kinematic equation:\[ v_f^2 = v_i^2 + 2a d \]with final velocity \( v_f = 0 \), initial velocity \( v_i = 22.4 \mathrm{m/s} \), and \( a = -3.50 \mathrm{m/s^2} \). Solving for \( d_3 \):\[ 0 = (22.4)^2 + 2(-3.50)d_3 \]\[ d_3 = \frac{(22.4)^2}{2 \times 3.50} = 71.68 \mathrm{m} \]
04

Sum distances to find total distance

Add up the distances for each phase to find the total distance covered by the train:\[ d_{total} = d_1 + d_2 + d_3 \]\[ d_{total} = 156.8 + 1568.0 + 71.68 = 1796.48 \mathrm{m} \]

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Key Concepts

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

Kinematic Equations
Kinematic equations are a set of formulas used to describe the motion of objects under uniform acceleration. These equations relate the five critical variables of motion: displacement (d), initial velocity (v_i), final velocity (v_f), acceleration (a), and time (t). In this exercise, we use kinematic equations to find the distances the train covers during different phases of its journey. An essential equation in our case was used to calculate the distance during acceleration:
  • \[ d_1 = \frac{1}{2}at^2 \]
This formula helps calculate the distance over time when starting from rest and moving under a constant acceleration. Knowing how to apply these equations is key to solving problems involving linear motion.
Constant Acceleration
Constant acceleration happens when an object's velocity changes at a consistent rate over time. In the context of the exercise, the train accelerates uniformly at 1.60 m/s虏 at the beginning of its journey. The importance of identifying this phase lies in applying the correct kinematic equation to compute the displacement during this period.
In real-world applications, constant acceleration is commonly observed in scenarios such as:
  • A car speeding up on a straight road.
  • A ball freely falling under gravity.
This uniform change in speed allows for straightforward calculations using kinematic equations.
Distance Calculation
Calculating distances in physics often involves applying kinematic equations that relate distance to time, acceleration, and velocity. In this problem, we found the total distance the subway train traveled by calculating distances for each phase of motion and adding them together. Each phase鈥攁cceleration, constant speed, and deceleration鈥攔equires its unique approach:
  • Acceleration: Use \[ d_1 = \frac{1}{2}at^2 \]
  • Constant Speed: Use \[ d_2 = v \times t \]
  • Deceleration: Solve \[ v_f^2 = v_i^2 + 2a d \] for \( d \)
These calculations help evaluate how far an object travels under various motion conditions. By understanding the method of distance calculation, students can solve complex problems systematically.
Velocity
Velocity is a vector quantity that tells us the speed and direction of an object's motion. It's crucial to distinguish velocity from speed, as speed is a scalar and only measures how fast something is moving, without regard to direction. During our train's journey:
  • The train starts at 0 m/s because it begins from rest.
  • It gains velocity as it accelerates to 22.4 m/s.
  • The train maintains this velocity during the constant speed phase.
  • Finally, it slows down, reaching a velocity of 0 m/s when stopping.
Velocity calculations greatly influence how we determine the distance and time of travel, offering detailed insights into an object's movement.

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Most popular questions from this chapter

You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 m/s passes you. When the rear of the bus is 12.0 m past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 m/s. 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?

The rocket-driven sled Sonic Wind No. \(2,\) used for investigating the physiological effects of large accelerations, runs on a straight, level track 1070 \(\mathrm{m}(3500 \mathrm{ft})\) long. Starting from rest, it can reach a speed of 224 \(\mathrm{m} / \mathrm{s}(500 \mathrm{m} \mathrm{i} / \mathrm{h})\) in 0.900 \(\mathrm{s}\) . (a) Compute the acceleration in \(\mathrm{m} / \mathrm{s}^{2},\) assuming that it is constant. (b) What is the ratio of this acceleration to that of a freely falling body \((g) ?\) (c) What distance is covered in 0.900 s? (d) A magazine article states that at the end of a certain run, the speed of the sled decreased from 283 \(\mathrm{m} / \mathrm{s}(632 \mathrm{mi} / \mathrm{h})\) to zero in 1.40 \(\mathrm{s}\) and that during this time the magnitude of the acceleration was greater than 40 \(\mathrm{g}\) . Are these figures consistent?

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 \(\mathrm{m}\) away at a constant speed of \(2.50 \mathrm{m} / \mathrm{s},\) returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?

Touchdown on the Moon. A lunar lander is making its descent to Moon Base I (Fig. 2.40). 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 \(\mathrm{m} / \mathrm{s}^{2}\) .

Certain rifles can fire a bullet with a speed of 965 \(\mathrm{m} / \mathrm{s}\) just as it leaves the muzzle (this speed is called the muzzle velocity). If the muzzle is 70.0 \(\mathrm{cm}\) long and if the bullet is accelerated uniformly from rest within it, (a) what is the acceleration (in \(g^{\prime}\) s) of the bullet in the muzzle, and (b) for how long (in ms) is it in the muzzle? (c) If, when this rifle is fired vertically, the bullet reaches a maximum height \(H,\) what would be the maximum height (in terms of \(H\) ) for a new rifle that produced half the muzzle velocity of this one?
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