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Chapter 2: Problem 81

A particle's acceleration along an \(x\) axis is \(a=5.0 t,\) with \(t\) in seconds and \(a\) in meters per second squared. At \(t=2.0 \mathrm{~s}\) its velocity is \(+17 \mathrm{~m} / \mathrm{s}\). What is its velocity at \(t=4.0 \mathrm{~s} ?\)

Short Answer

Expert verified
The velocity at \( t = 4 \, \text{seconds} \) is \( 47 \, \text{m/s} \).

Step by step solution

01

Identify Given Information

We are given the acceleration function of the particle, which is \( a(t) = 5.0t \). We know the velocity at \( t = 2.0 \, \text{seconds} \) is \( v(2) = +17 \, \text{m/s} \). We are tasked to find the velocity at \( t = 4.0 \, \text{seconds} \).
02

Set Up the Integral

Acceleration is the derivative of velocity with respect to time. To find the velocity function, we integrate the acceleration function \( a(t) = 5.0t \) with respect to time \( t \).
03

Perform the Integration

Integrate the acceleration function: \[ v(t) = \int a(t) \, dt = \int 5.0t \, dt \].This gives us: \[ v(t) = \frac{5.0}{2} t^2 + C, \] where \( C \) is the constant of integration.
04

Find the Constant of Integration

We use the initial condition \( v(2) = 17 \, \text{m/s} \) to find the constant. Substitute \( t = 2.0 \, \text{seconds} \) and \( v(2) = 17 \) into the velocity function:\[ 17 = \frac{5.0}{2}(2)^2 + C \]. Solve for \( C \):\[ 17 = 10 + C \]\[ C = 7 \].
05

Form the Complete Velocity Equation

Substitute the value of \( C \) back into the velocity function: \[ v(t) = \frac{5.0}{2} t^2 + 7 \].
06

Calculate Velocity at \( t = 4 \)

Use the velocity function to find the velocity at \( t = 4 \, \text{seconds} \):\[ v(4) = \frac{5.0}{2} (4)^2 + 7 \].Calculate \[ v(4) = \frac{5.0}{2} \times 16 + 7 \] . This simplifies to \( v(4) = 40 + 7 \), so \( v(4) = 47 \, \text{m/s} \).

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Key Concepts

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

Acceleration
In physics, acceleration refers to the rate at which an object changes its velocity. It tells us how quickly something is speeding up or slowing down. In our problem, the acceleration of the particle is described by the equation \( a(t) = 5.0t \). This equation means that the acceleration is not constant, but instead increases with time.

Understanding the given acceleration function reveals important information about the particle's motion.
  • Increasing acceleration: As time progresses, acceleration becomes larger, meaning the particle gains speed more rapidly.
  • Time-dependence: The fact that acceleration depends on time \( t \) indicates non-uniform motion.
Acceleration is a vector quantity, which means it has magnitude and direction. In this exercise, the acceleration is positive, suggesting that the particle is moving in a specific direction along the x-axis and is picking up speed.
Velocity
Velocity is a measure of how fast something is moving and in what direction. It differs from speed, as velocity includes the direction of movement.

In our task, we knew the velocity at \( t = 2.0 \, \text{seconds} \), which was \( +17 \, \text{m/s} \). We sought to find out the velocity at a later time (\( t = 4.0 \, \text{seconds} \)).
  • The acceleration function \( a(t) = 5.0t \) is critical because it shows us how the velocity of the particle changes over time.
  • To find the velocity at different instants, we derived the velocity function through integration. This function allowed us to compute the velocity at any given time.
Velocity plays a central role in motion analysis, providing insights into how the object's position changes over time. Remember, like acceleration, velocity is also a vector, characterized by magnitude and direction.
Integration
Integration is a mathematical method used to find a function when its rate of change is known. The connection between acceleration and velocity is that velocity is the integral of acceleration over time.

In our scenario, integrating the acceleration function \( a(t) = 5.0t \) helped us determine the velocity function. Here's how integration comes into play:
  • Setting up the integral: We integrated the acceleration function to find the general form of the velocity function, \( v(t) = \int 5.0t \, dt \).
  • Solving the integral: Performing the integration gave us \( v(t) = \frac{5.0}{2} t^2 + C \), where \( C \) is the integration constant.
The integration constant represents unknown initial conditions in scenarios like this. To resolve this, additional information like initial velocity is needed to solve for \( C \). Integration transforms the known rate of change (acceleration) into a tangible velocity function securely linked to the particle's motion.
Equations of Motion
Equations of motion describe the relationship between an object's velocity, acceleration, position, and time. In the context of this problem, we derived the velocity function from the acceleration function using these principles.

Here, we continue from our integration step to solve for velocity at specific times:
  • Using initial conditions: Knowing \( v(2) = 17 \, \text{m/s} \), we substituted \( v(t) \) to find the constant: \( C = 7 \).
  • Forming the velocity equation: With constant \( C \) determined, the velocity function became \( v(t) = \frac{5.0}{2} t^2 + 7 \).
  • Calculating at \( t = 4 \): Plugging into our equation gave us \( v(4) = 47 \, \text{m/s} \).
Equations of motion are vital in predicting and analyzing movement. They harmonize acceleration and velocity, enabling us to succinctly specify patterns of motion. By employing these equations, we derive rich insights into the story of moving particles and systems.

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Most popular questions from this chapter

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