Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that an object moves along a coordinate line so that its directed distance from the origin after \(t\) seconds is \(\sqrt{2 t+1}\) feet. (a) Find its instantaneous velocity at \(t=\alpha, \alpha>0\). (b) When will it reach a velocity of \(\frac{1}{2}\) foot per second? (see Example 5.)

Step-by-Step Solution

Answer

(a) Instantaneous velocity at \(t = \alpha\) is \((2\alpha + 1)^{-1/2}\). (b) It reaches a velocity of \(\frac{1}{2}\) ft/s at \(t = \frac{3}{2}\) seconds.

Step 1: Understand the Problem

The object's position as a function of time, \(t\), is given by \(s(t) = \sqrt{2t + 1}\) feet. We need to find the instantaneous velocity at any general time \(t = \alpha\), and determine when the velocity is \(\frac{1}{2}\) ft/s.

Step 2: Find the General Expression for Velocity

The velocity is the derivative of the position with respect to time. Differentiate \(s(t) = \sqrt{2t + 1}\) to get the velocity function. \[ v(t) = \frac{d}{dt}\left(\sqrt{2t + 1}\right) = \frac{1}{2}\cdot(2t+1)^{-1/2}\cdot2 = (2t+1)^{-1/2}. \]

Step 3: Compute the Instantaneous Velocity at \(t = \alpha\)

Substitute \(t = \alpha\) into the velocity function derived in Step 2. \[ v(\alpha) = (2\alpha + 1)^{-1/2}. \] This is the instantaneous velocity at \(t = \alpha\).

Step 4: Setup Equation for Specific Velocity

To find when the velocity equals \(\frac{1}{2}\), set \(v(t) = \frac{1}{2}\) and solve for \(t\).\[ (2t+1)^{-1/2} = \frac{1}{2}. \]

Step 5: Solve for \(t\) When Velocity is \(\frac{1}{2}\)

Squaring both sides of the equation from Step 4, \[ (2t+1)^{-1} = \frac{1}{4}. \]Invert both sides to simplify, \[ 2t+1 = 4. \]Solve for \(t\):\[ 2t = 3 \quad \Rightarrow \quad t = \frac{3}{2}. \] Therefore, the object reaches a velocity of \(\frac{1}{2}\) ft/s at \(t = \frac{3}{2}\) seconds.

Key Concepts

Derivative of Position

In calculus, the concept of derivative is crucial for understanding how functions change over time. The derivative of a position function, which describes the movement of an object, gives us the velocity function of that object. Essentially, the derivative tells us how fast the position of the object is changing at any given moment.
For the position function given in the problem, \[ s(t) = \sqrt{2t + 1}, \]we need to take the derivative with respect to time \(t\) to find the velocity function.
By employing the chain rule, we find:\[ v(t) = \frac{d}{dt}\left(\sqrt{2t + 1}\right) = (2t+1)^{-1/2}. \]
This expression represents the instantaneous rate of change of position, which is the instantaneous velocity.

Velocity Function

The velocity function is a mathematical expression that provides the velocity of an object at any given time. Velocity, unlike speed, is a vector quantity, meaning it has both magnitude and direction. For the object in the exercise, the velocity function is derived from the position function: \[ v(t) = (2t+1)^{-1/2}. \]
This function allows us to compute the object’s velocity at any particular moment \(t = \alpha\). Simply substitute the desired value of \(t\) to obtain the instantaneous velocity. For example, if we want to determine the velocity at \(t = \alpha\), we substitute \(\alpha\) into the velocity function:
\[ v(\alpha) = (2\alpha + 1)^{-1/2}. \]
This process illustrates how understanding derivatives underscores much of physics, providing insights into the behavior of moving objects.

Solving Differential Equations

Solving differential equations involves finding a function that satisfies a given relation involving derivatives. These equations are pivotal in understanding dynamic systems, such as the movement of an object along a path.
In our example, once we derive the velocity function from the position function, a differential equation arises that must be manipulated and solved to answer specific questions. For instance, to find the time at which velocity \(v(t) = \frac{1}{2}\), we set the equation:\[ (2t+1)^{-1/2} = \frac{1}{2}. \]
Solving this equation involves elevating each side to a power to eliminate the negative exponent:
\[ (2t+1)^{-1} = \frac{1}{4}, \]and subsequently solving for \(t\):
\[ 2t + 1 = 4. \]
This process highlights how differential equations are central to describing changes in physical systems.

Coordinate System

A coordinate system is the framework that allows us to specify the position of points in space. In physics, we often assume a simple one-dimensional coordinate system along a line or axis, where location is determined by a single variable, like \(t\) in our problem.
The object moves along this defined path, and each point in time corresponds uniquely to a position given by the position function. In this exercise, the coordinate system lets us describe the object’s distance from the origin as a function of time:\[ s(t) = \sqrt{2t + 1}. \]
This one-dimensional approach simplifies the calculation of velocity and other dynamic properties because it reduces complexity.
Understanding how to utilize coordinate systems allows us to organize information about an object's trajectory effectively, making it easier to apply calculus concepts like derivatives and integrals in solving physics problems.