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Chapter 2: Problem 2

Suppose \(s(t)\) is the position of an object moving along a line at time \(t \geq 0 .\) Describe a process for finding the instantaneous velocity at \(t=a\)

Short Answer

Expert verified
Answer: To find the instantaneous velocity, follow these steps: 1. Understand that the instantaneous velocity is the rate of change of the position, which can be found by differentiating the position function s(t) with respect to time. 2. Differentiate the position function s(t) to obtain the velocity function, v(t). 3. Evaluate the velocity function v(t) at t=a to find the instantaneous velocity at that specific time.

Step by step solution

01

Understand the instantaneous velocity concept

Instantaneous velocity is the rate at which an object's position is changing at any given time. This is described mathematically as the derivative of the position function with respect to time. In other words, we will need to find the derivative of the function s(t) concerning t.
02

Differentiate the position function

To calculate the derivative of the position function s(t), we will need to use differentiation rules, such as the power rule, product rule, or chain rule, if necessary. Once the position function s(t) is differentiated concerning t, we will obtain a new function representing the velocity of the object (v(t)).
03

Evaluate the velocity function at t=a

To find the instantaneous velocity at a specific time t=a, we need to evaluate the velocity function (obtained in step 2) at t=a. In other words, calculate v(a). This will give us the instantaneous velocity of the object at time t=a. The derivative of the position function concerning time is the key to finding the instantaneous velocity. Once we have the derivative, we evaluate it at the desired time to obtain the instantaneous velocity at that specific moment.

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Key Concepts

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

Derivative
The concept of a derivative is essential when dealing with functions. In simple terms, a derivative measures how a function changes as its input changes. Imagine you have a graph of a curve: the slope of the tangent line to that curve at any point is what the derivative represents. In the context of movement along a line, like tracking a car's motion, the derivative tells us how fast the position is changing. This idea becomes crucial when you want to find how quickly something moves at an exact moment in time.
Position function
The position function, noted typically as \(s(t)\), describes where an object is located along a line as a function of time \(t\). When an object moves, its position changes, and this is mathematically captured by this function. Knowing the position at different times allows us to understand the object's path. For example, if a car drives along a road, \(s(t)\) would tell us its exact location at any time \(t\). This groundwork is crucial for understanding how to compute the velocity by examining how this position changes.
Differentiation rules
Differentiation rules are the tools we use to find the derivative of more complex functions. When deriving the position function \(s(t)\), these rules come into play:
  • Power Rule: If \(f(t) = t^n\), then \(f'(t) = nt^{n-1}\).
  • Product Rule: If \(f(t) = u(t)v(t)\), then \(f'(t) = u'(t)v(t) + u(t)v'(t)\).
  • Chain Rule: If \(f(t) = g(h(t))\), then \(f'(t) = g'(h(t))h'(t)\).
These rules allow us to handle various forms of position functions. Whether they involve polynomials or complex multi-variable expressions, the rules guide us in simplifying the function to find its rate of change.
Velocity function
The velocity function, noted as \(v(t)\), is what you get after differentiating the position function \(s(t)\). It directly describes the rate of change of position with respect to time. It's like the speedometer of a moving object, telling us how fast the object is traveling at any point in time. With the velocity function, you can pinpoint the instantaneous velocity at a specific time \(t = a\). This is done by evaluating \(v(t)\) at \(t = a\), thus providing the speed at that precise moment. In essence, the velocity function is your roadmap to understanding speed changes over time.

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Most popular questions from this chapter

Sketch the graph of a function with the given properties. You do not need to find a formula for the function. $$\begin{array}{l} g(1)=0, g(2)=1, g(3)=-2, \lim _{x \rightarrow 2} g(x)=0 \\ \lim _{x \rightarrow 3^{-}} g(x)=-1, \lim _{x \rightarrow 3^{+}} g(x)=-2 \end{array}$$

Other techniques Evaluate the following limits, where a and \(b\) are fixed real numbers. $$\lim _{h \rightarrow 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}$$

Determine the end behavior of the following transcendental functions by evaluating appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist. $$f(x)=\sin x$$

Determine the following limits and justify your answers. $$\lim _{x \rightarrow 2} \sqrt{\frac{4 x+10}{2 x-2}}$$

Zero velocity A projectile is fired vertically upward and has a position given by \(s(t)=-16 t^{2}+128 t+192,\) for \(0 \leq t \leq 9\) a. Graph the position function, for \(0 \leq t \leq 9\) b. From the graph of the position function, identify the time at which the projectile has an instantaneous velocity of zero; call this time \(t=a\) c. Confirm your answer to part (b) by making a table of average velocities to approximate the instantaneous velocity at \(t=a\) d. For what values of \(t\) on the interval [0,9] is the instantaneous velocity positive (the projectile moves upward)? e. For what values of \(t\) on the interval [0,9] is the instantaneous velocity negative (the projectile moves downward)?
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