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Chapter 2: Problem 75

The acceleration of a particle is given by \(a_{x}(t)=\) \(-2.00 \mathrm{m} / \mathrm{s}^{2}+\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t .\) (a) Find the initial velocity \(v_{\mathrm{ax}}\) such that the particle will have the same \(x\) -coordinate at \(t=4.00 \mathrm{s}\) as it had at \(t=0\) (b) What will be the velocity at \(t=4.00 \mathrm{s} ?\)

Short Answer

Expert verified
(a) Initial velocity: \(-4.00 \, \mathrm{m/s}\) (b) Velocity at \(t=4.00 \mathrm{s}\): \(12.00 \, \mathrm{m/s}\)

Step by step solution

01

Integrate Acceleration to Find Velocity Function

The acceleration function is given by \(a_x(t) = -2.00 \ \mathrm{m/s^2} + 3.00 \ \mathrm{m/s^3} \cdot t\). To find the velocity function, integrate the acceleration with respect to time \(t\):\[v_x(t) = \int (-2.00 + 3.00t) \, dt = -2.00t + \frac{3.00}{2}t^2 + C\]where \(C\) is the constant of integration, representing the initial velocity \(v_{ax}\).
02

Integrate Velocity to Find Position Function

The velocity function derived is \(v_x(t) = -2.00t + \frac{3.00}{2}t^2 + C\). Integrate this function to find the position function:\[x(t) = \int (-2.00t + \frac{3.00}{2}t^2 + C) \, dt = -\frac{2.00}{2}t^2 + \frac{3.00}{6}t^3 + Ct + D\]Simplify:\[x(t) = -1.00t^2 + 0.50t^3 + Ct + D\]where \(D\) is another constant of integration representing the initial position.
03

Apply Initial Condition for Position

Given that the position at \(t=0\) is the same as at \(t=4.00\), we have \(x(0) = x(4.00)\). Substitute \(t = 0\) and \(t = 4.00\) into the position function:\[x(0) = 0 + 0 + C \cdot 0 + D = D\]\[x(4.00) = -1.00(4.00)^2 + 0.50(4.00)^3 + 4.00C + D\]Set \(x(0) = x(4.00)\):\[D = -16.00 + 32.00 + 4.00C + D\]Cancel \(D\):\[0 = 16.00 + 4.00C\]Solve for \(C\):\[C = -4.00\]This gives the initial velocity \(v_{ax} = -4.00 \, \mathrm{m/s}\).
04

Calculate Velocity at t=4.00 seconds

Use the velocity function derived in Step 1: \[v_x(t) = -2.00t + 1.50t^2 -4.00\]Substitute \(t = 4.00\) into this function:\[v_x(4.00) = -2.00(4.00) + 1.50(4.00)^2 - 4.00\]Calculate the value:\[v_x(4.00) = -8.00 + 24.00 - 4.00 = 12.00 \, \mathrm{m/s}\]Thus, the velocity at \(t=4.00\) seconds is \(12.00 \, \mathrm{m/s}\).

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Key Concepts

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

Acceleration Function
In particle kinematics, understanding the acceleration function is crucial as it describes how the velocity of a particle changes with time. For the problem, the acceleration function is expressed as
  • \(a_x(t) = -2.00 \, \mathrm{m/s^2} + 3.00 \, \mathrm{m/s^3} \cdot t\)
This formula indicates that the particle experiences a constant negative acceleration of \(-2.00 \, \mathrm{m/s^2}\), coupled with an increasing positive acceleration proportional to time \(t\) with a rate of \(3.00 \, \mathrm{m/s^3}\). Thus, as time advances, the positive component of the acceleration becomes more influential, potentially altering the particle's motion pattern significantly over time.
Velocity Integration
Velocity integration is a critical step for working with acceleration functions because it allows us to determine how velocity evolves over time given a specific acceleration. By integrating the acceleration function
  • \(-2.00 + 3.00t\)
with respect to \(t\), the velocity function becomes:
  • \(v_x(t) = -2.00t + \frac{3.00}{2}t^2 + C\)
Here, \(C\) is a constant that accounts for initial velocity, effectively capturing the knowledge about the velocity of the particle at time \(t = 0\). This integration establishes a relationship between the change in speed and time, allowing us to predict future motion based on initial conditions and applied accelerations.
Initial Conditions
Initial conditions are the values at the beginning of the observation, providing the foundation for solving kinematic problems. In this scenario, we know the particle's position at time \(t=0\) is the same as at \(t=4.00 \, \mathrm{s}\). These conditions help determine the unknown constants when integrating, such as when we set
  • \(x(0) = x(4.00)\)
From the position function
  • \(x(t) = -1.00t^2 + 0.50t^3 + Ct + D\)
we find \(C = -4.00\) by ensuring \(D\), the initial position, is consistent for both \(t=0\) and \(t=4.00\). Understanding and applying these conditions successfully connects both the initial and final states of motion, enabling precise calculations.
Position Function
The position function describes a particle's location at any given time, offering a visual and numerical tracking of its path. By integrating the velocity function
  • \(v_x(t) = -2.00t + \frac{3.00}{2}t^2 - 4.00\)
with respect to time, we derive the position function
  • \(x(t) = -1.00t^2 + 0.50t^3 -4.00t + D\)
Here, \(D\) is a constant of integration, representing the initial position. This function allows us to calculate where the particle is located at any point in time \(t\). Knowing how to manipulate and interpret the position function is critical for understanding particle kinematics and making predictions about future positions based on velocity and acceleration data.

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Most popular questions from this chapter

Entering the Freeway. A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 \(\mathrm{m} / \mathrm{s}(45 \mathrm{mi} / \mathrm{h})\) when it reaches the end of the 120 \(\mathrm{m}\) -long ramp. \((\mathrm{a})\) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 \(\mathrm{m} / \mathrm{s}\) . What distance does the traffic travel while the car is moving the length of the ramp?

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According to recent test data, an automobile travels 0.250 \(\mathrm{mi}\) in 19.9 \(\mathrm{s}\) s, starting from rest. The same car, when braking from 60.0 \(\mathrm{mi} / \mathrm{h}\) on dry pavement, stops in 146 \(\mathrm{ft}\) . Assume constant acceleration in each part of the motion, but not necessarily the same acceleration when slowing down as when speeding up. (a) Find the acceleration of this car when it is speeding up and when it is braking. \((b)\) If its acceleration is constant, how fast (in mi/h) should this car be traveling after 0.250 \(\mathrm{mi}\) of acceleration? The actual measured speed is 70.0 \(\mathrm{mi} / \mathrm{h}\) ; what does this tell you about the motion? (c) How long does it take this car to stop while braking from 60.0 \(\mathrm{mi} / \mathrm{h}\) ?

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