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In Exercises, \(s(t)\) is the position function of a body moving along a coordinate line; \(s(t)\) is measured in feet and \(t\) in seconds, where \(t \geq 0 .\) Find the position, velocity, and speed of the body at the indicated time. $$ s(t)=\frac{t}{t+1} ; \quad t=0 $$

Short Answer

Expert verified
At time \(t = 0\), the body is at position \(s(0) = 0\), with a velocity of \(s'(0) = 1\) and a speed of 1.

Step by step solution

01

Evaluate the Position Function at \(t = 0\)

To find the position of the body at $$t = 0$$, we substitute $$t = 0$$ into the position function $$s(t) = \frac{t}{t + 1}$$. Therefore, the position function at $$t = 0$$ is: $$ s(0) = \frac{0}{0 + 1} = \frac{0}{1} = 0 $$
02

Calculate the Derivative of the Position Function

To determine the derivative of the position function with respect to $$t$$, which represents the velocity, we must differentiate the function $$s(t) = \frac{t}{t + 1}$$. Using the quotient rule for differentiation, where if $$y = \frac{u}{v}$$, then $$y' = \frac{u'v - uv'}{v^2}$$, we have: $$ u = t \quad \Rightarrow \quad u' = 1 $$ $$ v = t + 1 \quad \Rightarrow \quad v' = 1 $$ Using this, we can find the derivative of $$s(t)$$: $$ s'(t) = \frac{1(t + 1) - t}{(t + 1)^2} = \frac{t + 1 - t}{(t + 1)^2} = \frac{1}{(t + 1)^2} $$
03

Evaluate the Derivative at \(t = 0\)

Evaluate the velocity function, $$s'(t) = \frac{1}{(t + 1)^2}$$, at $$t = 0$$ to obtain the velocity of the body at this time: $$ s'(0) = \frac{1}{(0 + 1)^2} = \frac{1}{1^2} = 1 $$
04

Calculate the Speed

Given that the speed is the absolute value of the velocity, and the velocity of the body at $$t = 0$$ is $$1$$, the speed at this time is: $$ \text{Speed} = \lvert s'(0) \rvert = \lvert1\rvert = 1 $$ In conclusion, at time $$t = 0$$, the body is at position $$0$$, with a velocity of $$1$$ and a speed of $$1$$.

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Key Concepts

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

Velocity
When studying motion, velocity refers to the change in the position of an object over time. In calculus, this is quantified as the derivative of the position function with respect to time. To put it simply, if you were to look at the position of a moving body at various times, velocity would tell you how fast its position is changing at any given moment.

For instance, with the position function
\(s(t) = \frac{t}{t+1}\), as given in our exercise, the velocity function is the derivative
\(s'(t) = \frac{1}{(t + 1)^2}\). This reveals that velocity is not just a measure of how fast the object is going but also the direction it is moving in, since velocity can be positive or negative. At
\(t = 0\), the velocity is \(1\) foot per second, indicating the object is moving in the positive direction at 1 foot per second at that instant.
Speed
Speed is often confused with velocity, but there is a crucial distinction: speed is the absolute magnitude of velocity, which means it does not take direction into account and is always a non-negative value. It tells you how fast an object is moving regardless of its traveling direction. In our exercise, since the velocity at
\(t = 0\) is \(1\) (calculated in the previous steps), the speed is simply the absolute value of this velocity, which is also \(1\) foot per second.

In general terms, if you have a velocity function
\(v(t)\), the speed at any time
\(t\) would be
\(|v(t)|\). This is particularly useful when you are only interested in how fast an object is going without being concerned about its direction of motion.
Quotient Rule Differentiation
The quotient rule is a technique for finding the derivative of a function that is the quotient of two other functions. It is especially handy when dealing with complex expressions where one function is divided by another. The formula is:
\[\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\] where \(u\) and \(v\) are functions of \(t\), and \(u'\) and \(v'\) are their respective derivatives with respect to \(t\).

In simpler terms, if you have a fraction and you need to find how it changes as \(t\) changes, you use the quotient rule to differentiate it. As applied to the exercise's position function
\(s(t) = \frac{t}{t+1}\), the quotient rule enables us to find the function's derivative (velocity), resulting in the simplified version
\(s'(t) = \frac{1}{(t + 1)^2}\), after applying the rule step by step.
Derivative of Position Function
The derivative of a position function with respect to time is a fundamental concept in calculus, particularly in the context of motion. It essentially represents the velocity of an object. If you have a function that tells you an object's location at any time, taking the derivative of that position function will give you a new function that tells you the object's velocity at any time.

Taking our exercise as an example, we started with the position function
\(s(t) = \frac{t}{t+1}\). Strictly speaking, the derivative
\(s'(t)\) of this function tells us how the position changes as time changes. By finding the derivative, we determined the velocity, which at the time
\(t = 0\) is \(1\) foot per second. This demonstrates how the derivative of the position function is a central tool in understanding motion.

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