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As a train accelerates away from a station, it reaches a specd of \(4.7 \mathrm{m} / \mathrm{s}\) in \(5.0 \mathrm{s}\). If the train's acceleration remains constant. what is its speed after an additional 6.0 s has elapsed?

Short Answer

Expert verified
The train's speed after an additional 6.0 seconds is 10.34 m/s.

Step by step solution

01

Understanding the Problem

We have a train that reaches a speed of 4.7 m/s in 5.0 seconds with a constant acceleration. We need to find the speed of the train after an additional 6.0 seconds from its current speed.
02

Calculate the Acceleration

Using the equation for constant acceleration, \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is time. We know \( v = 4.7 \mathrm{m/s} \), \( u = 0 \mathrm{m/s} \), and \( t = 5.0 \mathrm{s} \). Thus, \( 4.7 = 0 + a \cdot 5.0 \). Solve for \( a \): \( a = \frac{4.7}{5.0} = 0.94 \mathrm{m/s}^2 \).
03

Calculate the Speed after an Additional 6.0 s

Now that we know the acceleration is 0.94 m/s², we use the same formula for the subsequent time period: \( v = u + at \). Here, \( u = 4.7 \mathrm{m/s} \), \( a = 0.94 \mathrm{m/s}^2 \), and \( t = 6.0 \mathrm{s} \). Plug these values into the equation: \( v = 4.7 + 0.94 \cdot 6.0 \). This gives \( v = 4.7 + 5.64 = 10.34 \mathrm{m/s} \).
04

Review the Solution

The calculation confirms that after an additional 6.0 seconds, the speed of the train, given the constant acceleration, is 10.34 m/s.

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Key Concepts

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

Understanding Constant Acceleration
Constant acceleration occurs when an object's velocity changes at a steady rate over time. This means that, for each second that passes, the change in velocity is the same. Unlike variable acceleration, where the rate of velocity change differs at various moments, constant acceleration allows for simpler calculations using straightforward formulas.
In our train problem, the train accelerates uniformly from rest, indicating that its acceleration is constant. This is depicted in the calculation of the acceleration, which remains at 0.94 m/s² during the given time interval.
An acceleration system like this is linear, meaning that if we plot the velocity over time on a graph, the line would be straight. This linearity greatly simplifies predictions regarding velocity at future points.
Performing Velocity Calculations
Calculating velocity when acceleration is constant involves using the foundational physics formula: \[ v = u + at \] where:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration,
  • \( t \) is the time elapsed.
In our exercise, during the first 5.0 seconds, the train starts from rest, so its initial velocity \( u \) is 0. After calculating the acceleration as 0.94 m/s², the initial velocity is updated to 4.7 m/s.
For the additional 6.0 second period, this 4.7 m/s becomes the new initial velocity. Using the equation, we compute the final velocity after the additional time to find the total change in speed to be 10.34 m/s.
Exploring Equations of Motion
Equations of motion are mathematical tools that describe the behavior of objects in motion. They relate displacement, velocity, acceleration, and time. Typically, for constant acceleration, there are three primary equations:
  • \( v = u + at \) - to find final velocity.
  • \( s = ut + \frac{1}{2}at^2 \) - to calculate displacement.
  • \( v^2 = u^2 + 2as \) - to connect velocity and displacement without time.
For the train scenario, we utilized the first equation to predict the train's future speed after a known time period had passed.
This family of equations is highly versatile. They allow us to switch between finding speed, the distance covered, or how quickly an object is speeding up. Understanding when to apply each equation is made easier with practice and familiarity.

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Most popular questions from this chapter

Running with an initial velocity of \(+11 \mathrm{m} / \mathrm{s},\) a horse has an average acceleration of \(-1.81 \mathrm{m} / \mathrm{s}^{2}\). How long does it take for the horse to decrease its velocity to \(+6.5 \mathrm{m} / \mathrm{s}\) ?

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