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Chapter 4: Problem 28

A car with velocity \(2.00 \mathrm{~m} / \mathrm{s}\) at \(t=0\) accelerates at \(4.00 \mathrm{~m} / \mathrm{s}^{2}\) for \(2.50 \mathrm{~s}\). What is its velocity at \(t=2.50 \mathrm{~s}\) ?

Short Answer

Expert verified
The car's velocity at \( t = 2.50 \, \text{s} \) is \( 12.00 \, \text{m/s} \).

Step by step solution

01

Identify Initial Velocity

Label the initial velocity of the car as \( u \). According to the problem, the car's initial velocity is \( u = 2.00 \, \text{m/s} \).
02

Identify Acceleration and Time

Label the acceleration of the car as \( a \). According to the problem, the car accelerates at \( a = 4.00 \, \text{m/s}^2 \). The time period given is \( t = 2.50 \, \text{s} \).
03

Use the First Equation of Motion

The first equation of motion is \( v = u + at \), where \( v \) is the final velocity. We will use this equation to find the car's velocity at \( t = 2.50 \, \text{s} \).
04

Plug in the Known Values

Replace \( u \), \( a \), and \( t \) in the equation with the given values: \( v = 2.00 + 4.00 \times 2.50 \).
05

Perform the Calculations

Calculate \( 4.00 \times 2.50 = 10.00 \), then add \( 2.00 + 10.00 \) to get \( v = 12.00 \, \text{m/s} \).
06

State the Final Velocity

The car's velocity at \( t = 2.50 \, \text{s} \) is \( 12.00 \, \text{m/s} \).

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Key Concepts

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

Understanding Initial Velocity
In physics, initial velocity (\( u \)) is the velocity of an object at the start of a time interval. It is a crucial concept in kinematics because it serves as the starting point in analyzing an object's motion. Initial velocity is often given in problems explicitly. In real-world scenarios, you may need to determine it based on context or other given data.

For example, in our exercise, the car's initial velocity is \( 2.00 \, \text{m/s} \).
Understanding this helps us predict how the car's speed will change over time with the influence of additional forces like acceleration.
What is Acceleration?
Acceleration (\( a \)) refers to how quickly an object's velocity changes over time. It is a vector quantity, meaning it includes both magnitude and direction. Acceleration can be positive, negative, or zero, depending on whether an object speeds up, slows down, or moves at a constant velocity.
  • A positive acceleration means the object is increasing its speed.
  • Negative acceleration, often called deceleration, means the object is slowing down.
  • Zero acceleration indicates constant velocity.

In our problem, the car accelerates at \( 4.00 \, \text{m/s}^2 \), meaning it speeds up by \( 4.00 \, \text{m/s} \) every second.
First Equation of Motion Explained
The first equation of motion is a fundamental formula in kinematics that relates initial velocity, acceleration, and time to the final velocity of an object. It is given by the equation: \( v = u + at \). In this formula:
  • \( v \) is the final velocity.
  • \( u \) is the initial velocity.
  • \( a \) is the acceleration.
  • \( t \) is the time duration of the motion.

This equation helps us calculate how an object's velocity changes with respect to time when subjected to a constant acceleration.
It is particularly useful in scenarios where you have consistent acceleration and need to find out how fast an object will be moving after a certain period.
Calculating Final Velocity
The final velocity (\( v \)) of an object can be determined using the first equation of motion when initial velocity, acceleration, and time are known.
In the provided problem, the equation: \( v = 2.00 \, \text{m/s} + (4.00 \, \text{m/s}^2 \times 2.50 \, \text{s}) \) is used.

Perform the multiplication first: \( 4.00 \times 2.50 = 10.00 \).
Then add the product to the initial velocity: \( 2.00 + 10.00 = 12.00 \).

Thus, the car's velocity at \( 2.50 \) seconds is \( 12.00 \, \text{m/s} \).
This step illustrates how calculations based on equations of motion can predict velocities in various physical contexts.

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Most popular questions from this chapter

Find the average speed (in the given units) of an auto that travels each distance in the given time \(785 \mathrm{ft}\) in \(11.5 \mathrm{~s}\) (in \(\mathrm{ft} / \mathrm{s}\) )

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How long does it take a rock to drop \(95.0 \mathrm{~m}\) from rest? Find the final speed of the rock.
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