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Chapter 2: Problem 41

A locomotive is accelerating at \(1.6 \mathrm{m} / \mathrm{s}^{2} .\) It passes through a \(20.0-\mathrm{m}-\) wide erossing in a time of \(2.4 \mathrm{s}\). After the locomotive leaves the crossing, how much time is required until its speed reaches \(32 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
15.29 seconds.

Step by step solution

01

Identify Initial Conditions

The problem provides that the locomotive is accelerating at \(1.6 \, \text{m/s}^2\). It passes a \(20.0 \, \text{m}\) wide crossing in \(2.4 \, \text{s}\). We need to find the time required for the speed to reach \(32 \, \text{m/s}\).
02

Calculate Initial Velocity

Using the formula for displacement under constant acceleration: \(s = ut + \frac{1}{2}at^2\), where \(s\) is the displacement, \(u\) is the initial velocity, \(a\) is acceleration, and \(t\) is time. We plug in \(s = 20.0 \, \text{m}, a = 1.6 \, \text{m/s}^2, t = 2.4 \, \text{s}\) to solve the equation: \[20.0 = u(2.4) + \frac{1}{2}(1.6)(2.4)^2\]Solving for \(u\), we find \(u = 3.5 \, \text{m/s}\).
03

Calculate Final Velocity in the Crossing

Using the formula for final velocity: \(v = u + at\), where \(v\) is the final velocity after the crossing, we can substitute \(u = 3.5 \, \text{m/s}, a = 1.6 \, \text{m/s}^2, t = 2.4 \, \text{s}\).\[v = 3.5 + (1.6)(2.4)\]This gives \(v = 7.34 \, \text{m/s}\).
04

Determine Time to Reach Target Speed

Next, determine the time needed to accelerate from the final speed after the crossing, \(v = 7.34 \, \text{m/s}\), to the target speed, \(v = 32 \, \text{m/s}\). Using the formula for time: \( t = \frac{v - u}{a} \),\[t = \frac{32 - 7.34}{1.6}\]Thus, \(t = 15.29 \, \text{s}\).

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause the motion. It helps us to understand how things move by using mathematical formulas and concepts. In kinematics, we often deal with quantities such as velocity, acceleration, time, and displacement. These elements help us calculate and predict the motion of objects such as trains, cars, and other moving entities.
Key kinematic quantities are:
  • Displacement: The change in position of an object. It is a vector quantity.
  • Velocity: The rate of change of displacement, which can be initial or final.
  • Acceleration: The rate of change of velocity, which can be positive or negative.
  • Time: The duration over which motion occurs.
In our exercise, the locomotive's motion through a crossing involves constant acceleration, where kinematic equations are used to determine initial and final velocities and time durations.
Constant Acceleration Formulas
When an object moves with a constant acceleration, specific equations are used to describe its motion. These equations are foundational in solving physics problems related to motion.
A commonly used formula in constant acceleration problems is:
    • Displacement formula: \( s = ut + \frac{1}{2} a t^2 \)
    • Final velocity formula: \( v = u + at \)
    Here, \(s\) represents displacement, \(u\) is the initial velocity, \(v\) is the final velocity, \(a\) is the constant acceleration, and \(t\) is the time.Using these equations, one can solve for unknown variables, like in our problem where we calculated the initial and final velocities of the locomotive as it passes through a crossing while accelerating at \(1.6 \text{ m/s}^2\).
Initial and Final Velocity Calculations
Understanding initial and final velocities is crucial in addressing many kinematic problems. The initial velocity is the speed of an object before a given process starts, while the final velocity is the speed after the process.
In our exercise, we determined these velocities using the constant acceleration and the time taken across the crossing:
  • To find the initial velocity, we rearranged the displacement formula: \( s = ut + \frac{1}{2}at^2 \).
  • For the step-by-step calculation: Solving \( 20.0 = u(2.4) + \frac{1}{2}(1.6)(2.4)^2 \) led us to find \( u = 3.5 \text{ m/s} \).
  • For the final velocity, the formula used was \( v = u + at \), giving us \( v = 7.34 \text{ m/s} \).
These calculations are essential to determine the subsequent steps in solving the physics problem, such as the time required to reach a certain speed.
Time of Acceleration
The time of acceleration refers to the duration it takes for an object under constant acceleration to reach a designated speed. In many problems, including our locomotive scenario, this requires understanding the change in velocity and the constant acceleration.
We calculated the time it would take for the locomotive to accelerate from its final speed after leaving the crossing to reach its target speed.
  • Using the formula: Time is found by \( t = \frac{v - u}{a} \), where \( v \) is the target or final velocity, \( u \) is the initial velocity at the point considered, and \( a \) is the acceleration.
  • In the problem, the time to increase from \( 7.34 \text{ m/s} \) to \( 32 \text{ m/s} \) was determined by substituting \( a = 1.6 \text{ m/s}^2 \). This resulted in \( t = 15.29 \text{ s} \).
Such calculations help predict how long it will take for moving objects to reach desired speeds.

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Most popular questions from this chapter

The Kentucky Derby is held at the Churchill Downs track in Louisville, Kentucky. The track is one and one-quarter miles in length. One of the most famous horses to win this event was Secretariat. In 1973 he set a Derby record that would be hard to beat. His average acceleration during the last four quarter-miles of the race was \(+0.0105 \mathrm{m} / \mathrm{s}^{2}\). His velocity at the start of the final mile \((x=+1609 \mathrm{m})\) was about \(+16.58 \mathrm{m} / \mathrm{s}\). The acceleration, although small, was very important to his victory. To assess its effect, determine the difference between the time he would have taken to run the final mile at a constant velocity of \(+16.58 \mathrm{m} / \mathrm{s}\) and the time he actually took. Although the track is oval in shape, assume it is straight for the purpose of this problem.

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board), assuming that the dowmward direction is chosen as the negative direction./ (b) What is the highest point he reaches above the water?

One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is \(1.50 \mathrm{km}\). They start at the west side of the lake and head due south to begin with. (a) What is the distance they travel? (b) What are the magnitude and direction (relative to due east) of the couple's displacement?

Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, cycle A has an average acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) due east, while cycle \(\overrightarrow{\mathrm{B}}\) has an average acceleration of \(4.0 \mathrm{m} / \mathrm{s}^{2}\) due east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster?

Two soccer players start from rest, \(48 \mathrm{m}\) apart. They run directly toward each other, both players accelerating. The first player's acceleration has a magnitude of \(0.50 \mathrm{m} / \mathrm{s}^{2}\). The second player's acceleration has a magnitude of \(0.30 \mathrm{m} / \mathrm{s}^{2}\). (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?
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