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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 \( t = 4.0 \) s is \( 47 \) m/s.

Step by step solution

01

Understand the Given Problem

We are given an acceleration function \( a = 5.0t \), where \( a \) is in meters per second squared and \( t \) is in seconds. We need to find the velocity at \( t = 4.0 \) seconds, given that the velocity at \( t = 2.0 \) seconds is \( +17 \) m/s.
02

Integrate Acceleration to Find Velocity Equation

Since acceleration is the derivative of velocity, we integrate the acceleration function \( a(t) = 5.0t \) with respect to time to find the velocity function. This gives\[ v(t) = \int 5.0 t \, dt = 5.0 \cdot \frac{t^2}{2} + C = 2.5t^2 + C\]where \( C \) is the integration constant.
03

Solve for Integration Constant Using Initial Condition

We are given that the velocity \( v = 17 \) m/s when \( t = 2.0 \) s. Substitute these values into the velocity equation:\[ 17 = 2.5(2.0)^2 + C\]Solving for \( C \), we find:\[ 17 = 2.5 \cdot 4 + C \Rightarrow 17 = 10 + C \Rightarrow C = 7\]So, the velocity function becomes \( v(t) = 2.5t^2 + 7 \).
04

Calculate Velocity at t = 4.0 s

Now, use the velocity equation to find the velocity at \( t = 4.0 \) s:\[ v(4.0) = 2.5(4.0)^2 + 7\]Calculating this gives:\[ v(4.0) = 2.5 \cdot 16 + 7 = 40 + 7 = 47 \text{ m/s}\]Thus, the velocity at \( t = 4.0 \) s is \( 47 \) 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 is a fundamental concept that describes how quickly an object's velocity changes over time. Acceleration is usually given in meters per second squared (m/s虏), indicating how much the velocity increases per second. For this exercise, the acceleration of the particle is provided as a function of time, \( a(t) = 5.0t \). This tells us that acceleration is not constant; it increases linearly with time.
  • The factor of 5.0 indicates how fast the acceleration increases each second.
  • If \( t = 0 \), the acceleration is zero, meaning the particle starts from a rest or constant velocity situation.
  • As \( t \) increases, the acceleration grows, causing the particle to speed up more rapidly.
Understanding how acceleration affects motion helps us to predict how the velocity will change with time, which is essential in solving our problem.
Velocity
Velocity describes the speed and direction of a moving object. It is a vector quantity, which means it involves both magnitude and direction. In this exercise, we're dealing with velocity on a single axis, so we'll focus on its magnitude, measured in meters per second (m/s). To find the velocity at a specific time, we must first determine a function that describes how velocity changes over time, starting from the given acceleration function.
  • The given velocity at \( t = 2.0 \) seconds is \( +17 \text{ m/s} \), serving as an essential initial condition.
  • Velocity can increase or decrease, depending on whether the acceleration is positive or negative.
  • In our case, the acceleration is positive, suggesting that the velocity will rise as time progresses.
Once we create the velocity function \( v(t) = 2.5t^2 + 7 \), it allows us to compute the velocity at any time, including \( t = 4.0 \) seconds.
Integration
Integration is a powerful mathematical tool that allows us to find functions when we know their rates of change. Here, we integrate the given acceleration function to derive the velocity function. Since acceleration is the derivative of velocity with respect to time, integrating acceleration gives us the velocity.The mathematical process goes like this:\[v(t) = \int a(t) \, dt = \int 5.0t \, dt = 2.5t^2 + C\]where \( C \) is the constant of integration which represents the initial value of velocity.
  • Integration can be thought of as summing up all the tiny slices of acceleration over time.
  • The result provides a function that smoothly represents velocity over time, accounting for the changing acceleration.
  • Don't forget to include the integration constant! It captures the initial conditions that we know.
Initial Conditions
Initial conditions are like the starting point for equations that contain unknown constants. They allow us to solve for these constants and complete the function. In this problem, the initial condition is given as \( v = 17 \text{ m/s} \) when \( t = 2.0 \) seconds.By substituting this condition into the equation derived from integration, we can solve for the integration constant \( C \).
  • Substitute \( t = 2.0 \) and \( v(t) = 17 \) into the integrated formula: \[ 17 = 2.5(2.0)^2 + C \]
  • Solve for \( C \) which helps finalize the velocity function: \( C = 7 \).
  • Including initial conditions ensures that the function accurately reflects the situation right from the start.
Without considering initial conditions, the results could miss key aspects of the problem, leading to incorrect solutions.

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Most popular questions from this chapter

A stone is dropped into a river from a bridge \(43.9 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. The stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.

A motorcyclist who is moving along an \(x\) axis directed toward the east has an acceleration given by \(a=(6.1-1.2 t) \mathrm{m} / \mathrm{s}^{2}\) for \(0 \leq t \leq 6.0 \mathrm{~s}\). At \(t=0\), the velocity and position of the cyclist are \(2.7 \mathrm{~m} / \mathrm{s}\) and \(7.3 \mathrm{~m} .\) (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel between \(t=0\) and \(6.0 \mathrm{~s} ?\)

A car travels up a hill at a constant speed of \(40 \mathrm{~km} / \mathrm{h}\) and returns down the hill at a constant speed of \(60 \mathrm{~km} / \mathrm{h}\). Calculate the average speed for the round trip.

Two diamonds begin a free fall from rest from the same height, \(1.0 \mathrm{~s}\) apart. How long after the first diamond begins to fall will the two diamonds be \(10 \mathrm{~m}\) apart?

A train started from rest and moved with constant acceleration. At one time it was traveling \(30 \mathrm{~m} / \mathrm{s}\), and \(160 \mathrm{~m}\) farther on it was traveling \(50 \mathrm{~m} / \mathrm{s}\). Calculate (a) the acceleration, \((\mathrm{b})\) the time re- quired to travel the \(160 \mathrm{~m}\) mentioned, \((\mathrm{c})\) the time required to attain the speed of \(30 \mathrm{~m} / \mathrm{s}\), and (d) the distance moved from rest to the time the train had a speed of \(30 \mathrm{~m} / \mathrm{s}\). (e) Graph \(x\) versus \(t\) and \(y\) versus \(t\) for the train, from rest.
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