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Chapter 3: Problem 16

An elevator is designed to start from a resting position without a jerk. It can do this if the acceleration function is continuous. a. Show that the acceleration is continuous at \(t=0\) for the following position function \(s(t)=\left\\{\begin{aligned} 0, & \text { if } t<0 \\\ t^{3} &, \text { if } t \geq 0 \end{aligned}\right.\) b. What happens to the velocity and acceleration for very large values of \(t ?\)

Short Answer

Expert verified
The acceleration is continuous at \(t=0\), and both velocity and acceleration tend to infinity for large \(t\).

Step by step solution

01

Find the velocity function

To determine if the acceleration is continuous, we first need to find the velocity function. The velocity function is the derivative of the position function \(s(t)\).\For \(t < 0\), \(s(t) = 0\). So, \(v(t) = \frac{d}{dt}[0] = 0\).\For \(t \geq 0\), \(s(t) = t^3\). Therefore, \(v(t) = \frac{d}{dt}[t^3] = 3t^2\).\The velocity function is thus \(v(t) = \left\{\begin{aligned} 0, & \text{ if } t < 0 \ 3t^2, & \text{ if } t \geq 0 \end{aligned}\right.\).
02

Find the acceleration function

Now find the derivative of the velocity function to get the acceleration function.\For \(t < 0\), \(v(t) = 0\) so \(a(t) = \frac{d}{dt}[0] = 0\).\For \(t \geq 0\), \(v(t) = 3t^2\) so \(a(t) = \frac{d}{dt}[3t^2] = 6t\).\Thus, the acceleration function is \(a(t) = \left\{\begin{aligned} 0, & \text{ if } t < 0 \ 6t, & \text{ if } t \geq 0 \end{aligned}\right.\).
03

Check continuity of acceleration at \(t=0\)

The acceleration is continuous at \(t=0\) if the left-hand limit and right-hand limit of \(a(t)\) at \(t=0\) are equal and \(a(0)\) exists.\- Left-hand limit as \(t\to 0^-\) is \(a(0^-) = 0\).\- Right-hand limit as \(t\to 0^+\) is \(a(0^+) = 6 \cdot 0 = 0\).\- \(a(0) = 6 \cdot 0 = 0\).\Since \(a(0^-) = a(0^+) = a(0) = 0\), the acceleration is continuous at \(t=0\).
04

Analyze velocity and acceleration for large \(t\) values

For very large values of \(t\), the velocity \(v(t) = 3t^2\) will also become very large, as it grows with the square of \(t\).\The acceleration \(a(t) = 6t\) likewise becomes very large, increasing linearly with \(t\).\Thus, both velocity and acceleration tend to infinity as \(t\) increases without bound.

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Key Concepts

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

Position Function
In the context of motion, a position function describes the location of an object at any given time. It's usually represented by the function \( s(t) \), where \( t \) indicates time and \( s(t) \) represents the position of the object at that time.
In our problem, the position function is defined in pieces:
  • For \( t < 0 \), \( s(t) = 0 \). This means if we consider time before zero, the position of the elevator does not change and remains at the origin (zero position).
  • For \( t \geq 0 \), \( s(t) = t^3 \). Here, the position changes with the cube of the time, which indicates a smooth curve because cubing a value always gives a smooth increase without jumps or abrupt changes.
Understanding the position function is crucial because it serves as a foundation to calculate velocity and acceleration, which are derivatives of this function.
Velocity Function
Velocity is the rate at which an object's position changes over time. It's the derivative of the position function with respect to time, which reveals how fast the object is moving at any given point in time.
Let's dissect the velocity function in this case:
  • For \( t < 0 \), since the position \( s(t) = 0 \), the velocity \( v(t) = \frac{d}{dt}[0] = 0 \). This indicates there is no movement before time zero, the object remains stationary.
  • For \( t \geq 0 \), the velocity \( v(t) = \frac{d}{dt}[t^3] = 3t^2 \). This result shows that velocity increases as the square of the time. As \( t \) grows, \( 3t^2 \) becomes larger, meaning the elevator speeds up progressively.
Through this function, we see how the rate of change of position evolves and sets the stage for analyzing acceleration, which is the rate of change of this velocity function.
Acceleration Function
Acceleration reflects how quickly velocity itself changes with time. It is the derivative of the velocity function. By examining it, we understand the rate at which the object's speed increases or decreases.
The acceleration function in this case is:
  • For \( t < 0 \), because \( v(t) = 0 \), the acceleration \( a(t) = \frac{d}{dt}[0] = 0 \). Since there's no initial movement or velocity change outside \( t=0 \), there's no acceleration either.
  • For \( t \geq 0 \), \( a(t) = \frac{d}{dt}[3t^2] = 6t \). This shows the acceleration linearly increases with time, meaning the speed of the elevator is increasing steadily with time.
What's essential here is ensuring the continuity of acceleration at \( t=0 \). We verified the acceleration's left-hand and right-hand limits at \( t=0 \) are equal to the value of acceleration at \( t=0 \). This confirms that the acceleration function is continuous, fulfilling the requirement for a smooth start without a jerk.

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Most popular questions from this chapter

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