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A train on a straight, level track has an initial speed of \(35.0 \mathrm{~km} / \mathrm{h}\). A uniform acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) is applied while the train travels \(200 \mathrm{~m}\). (a) What is the speed of the train at the end of this distance? (b) How long did it take for the train to travel the \(200 \mathrm{~m} ?\)

Short Answer

Expert verified
The final speed is 26.35 m/s, and the time taken is 11.09 seconds.

Step by step solution

01

Convert Initial Speed

Convert the initial speed from km/h to m/s.Speed conversion formula: \[1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\]So the initial speed \(v_i = 35.0\text{ km/h} \times \frac{1}{3.6} \approx 9.72 \text{ m/s} \).
02

Identify Known Variables for Part (a)

We know the following: - Initial speed, \(v_i = 9.72 \text{ m/s}\)- Acceleration, \(a = 1.50 \text{ m/s}^2\)- Distance, \(s = 200 \text{ m}\)
03

Use Kinematic Equation to Find Final Speed

Use the kinematic equation: \[v_f^2 = v_i^2 + 2 a s\]Substitute the known values: \[v_f^2 = (9.72)^2 + 2 \times 1.50 \times 200\]Calculate to find \(v_f\).
04

Calculate Final Speed

Calculate the expression:\[v_f^2 = 94.4784 + 600\]\[v_f^2 = 694.4784\]\[v_f = \sqrt{694.4784} \approx 26.35 \text{ m/s}\]Thus, the final speed of the train is approximately \(26.35 \text{ m/s}\).
05

Identify Known Variables for Part (b)

Use the values:- Initial speed, \(v_i = 9.72 \text{ m/s}\)- Final speed, \(v_f = 26.35 \text{ m/s}\)- Acceleration, \(a = 1.50 \text{ m/s}^2\)
06

Use Kinematic Equation to Find Time

Use the equation: \[v_f = v_i + a t\]Solve for \(t\): \[t = \frac{v_f - v_i}{a} = \frac{26.35 - 9.72}{1.50}\]
07

Calculate Time

Calculate the time:\[t = \frac{16.63}{1.50} \approx 11.09 \text{ seconds}\]Thus, it takes approximately \(11.09\) seconds for the train to travel 200 meters.

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Key Concepts

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

Uniform Acceleration
Uniform acceleration refers to a state where an object speeds up or slows down at a consistent rate over time. This means that the object's velocity changes by the same amount in every equal time period.
For example, in our exercise, the train experiences a uniform acceleration of 1.50 m/s². This acceleration occurs in a straight and predictable manner along its path.
Understanding uniform acceleration is crucial for solving motion problems because it allows you to use specific kinematic equations. These equations help find other motion parameters, such as final speed or travel time, if the acceleration is constant.
Speed Conversion
In order to solve problems involving speed in physics, converting units accurately is a vital step. Many problems present speeds in kilometers per hour (km/h), but for consistency, especially in equations involving acceleration in meters per second squared (m/s²), speeds are converted to meters per second (m/s).
The conversion factor between km/h to m/s is given by:
  • 1 km/h = \( \frac{1}{3.6} \text{ m/s} \)
  • Therefore, for a speed of 35 km/h, the conversion is: \( 35 \times \frac{1}{3.6} \approx 9.72 \text{ m/s} \)
    This conversion is essential to ensure that all units are compatible, making calculations straightforward and error-free.
Kinematic Equations
Kinematic equations are core tools in physics for analyzing motion, especially when dealing with constant acceleration. Each equation relates different aspects of motion, such as initial velocity, final velocity, acceleration, time, and displacement.
A commonly used kinematic equation is:
  • \[ v_f^2 = v_i^2 + 2as \]
Here, "\(v_f\)" is the final velocity, "\(v_i\)" is the initial velocity, "\(a\)" is acceleration, and "\(s\)" is displacement.
This equation allows you to find the final speed of the train after a known distance, knowing its initial speed and the uniform acceleration.
Distance Calculation
Distance calculation in terms of kinematic motion refers to determining how far an object travels under a specific set of conditions. It often involves variables like speed, time, and acceleration.
When dealing with uniform acceleration, distance can also be part of the equation to find other variables. For instance, in our problem, given the initial speed, final speed, and acceleration, you can verify the distance traveled using the kinematic equations.
Accurately calculating distance aids in verifying findings or setting initial conditions for solving physics problems effectively. Keeping all calculations consistent with units ensures integrity and correctness in problem-solving.

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Most popular questions from this chapter

A tennis ball is dropped from a height of \(10.0 \mathrm{~m}\). It rebounds off the floor and comes up to a height of only \(4.00 \mathrm{~m}\) on its first rebound. (Ignore the small amount of time the ball is in contact with the floor.) (a) Determine the ball's speed just before it hits the floor on the way down. (b) Determine the ball's speed as it leaves the floor on its way up to its first rebound height. (c) How long is the ball in the air from the time it is dropped until the time it reaches its maximum height on the first rebound?

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