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Chapter 3: Problem 3

Each of the following position functions describes the motion of an object along a straight line. Find the velocity and acceleration as functions of \(t, t \geq 0\) a. \(s(t)=5 t^{2}-3 t+15\) b. \(s(t)=2 t^{3}+36 t-10\) c. \(s(t)=t-8+\frac{6}{t}\) d. \(s(t)=(t-3)^{2}\) e. \(s(t)=\sqrt{t+1}\) f. \(s(t)=\frac{9 t}{t+3}\)

Short Answer

Expert verified
a: \(v(t) = 10t - 3\), \(a(t) = 10\); b: \(v(t) = 6t^2 + 36\), \(a(t) = 12t\); c: \(v(t) = 1 - \frac{6}{t^2}\), \(a(t) = \frac{12}{t^3}\); d: \(v(t) = 2(t-3)\), \(a(t) = 2\); e: \(v(t) = \frac{1}{2\sqrt{t+1}}\), \(a(t) = -\frac{1}{4(t+1)^{3/2}}\); f: \(v(t) = \frac{27}{(t+3)^2}\), \(a(t) = -\frac{54}{(t+3)^3}\).

Step by step solution

01

Calculate Velocity for Function a

The position function is given by \(s(t) = 5t^2 - 3t + 15\). The velocity is the first derivative of the position function, \(v(t) = s'(t)\). Differentiate to find \(v(t) = 10t - 3\).
02

Calculate Acceleration for Function a

Now, find the acceleration by taking the derivative of the velocity function. Differentiate \(v(t) = 10t - 3\) to get \(a(t) = 10\).
03

Calculate Velocity for Function b

For \(s(t) = 2t^3 + 36t - 10\), find \(v(t)\) by differentiating the position function: \(v(t) = 6t^2 + 36\).
04

Calculate Acceleration for Function b

Differentiate \(v(t) = 6t^2 + 36\) to find \(a(t) = 12t\).
05

Calculate Velocity for Function c

The position function is \(s(t) = t - 8 + \frac{6}{t}\). Differentiate to find the velocity: \(v(t) = 1 - \frac{6}{t^2}\).
06

Calculate Acceleration for Function c

Differentiate \(v(t) = 1 - \frac{6}{t^2}\) to get \(a(t) = \frac{12}{t^3}\).
07

Calculate Velocity for Function d

Given \(s(t) = (t-3)^2\), differentiate to find \(v(t) = 2(t-3)\).
08

Calculate Acceleration for Function d

Differentiate \(v(t) = 2(t-3)\) to find \(a(t) = 2\).
09

Calculate Velocity for Function e

The position function is \(s(t) = \sqrt{t+1}\). The velocity is \(v(t) = \frac{1}{2\sqrt{t+1}}\) after differentiation.
10

Calculate Acceleration for Function e

Differentiate \(v(t) = \frac{1}{2\sqrt{t+1}}\) to find \(a(t) = -\frac{1}{4(t+1)^{3/2}}\).
11

Calculate Velocity for Function f

Given \(s(t) = \frac{9t}{t+3}\), use the quotient rule to find \(v(t) = \frac{27}{(t+3)^2}\).
12

Calculate Acceleration for Function f

Differentiate \(v(t) = \frac{27}{(t+3)^2}\) using the chain rule to find \(a(t) = -\frac{54}{(t+3)^3}\).

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Key Concepts

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

Velocity Function
The velocity of an object in motion along a straight line can be understood as how fast the object changes its position over time. Mathematically, the velocity function, denoted as \(v(t)\), represents the first derivative of the position function \(s(t)\). This means we're finding how quickly the position \(s(t)\) is changing with respect to time \(t\). By taking the derivative, if \(s(t) = 5t^2 - 3t + 15\), the velocity function becomes \(v(t) = 10t - 3\). This tells us the rate at which the object is moving at any time \(t\).In calculus, finding the velocity function is the initial step to understanding the motion characteristics of an object. It's much like determining the speed of a car at different moments during a trip. To find the velocity function for any position function:
  • Differentiate the position function with respect to time \(t\).
  • The result gives the velocity function \(v(t)\).
This differentiation step allows us to peek into the instantaneous rate of change of the object's position.
Acceleration Function
Acceleration tells us how the velocity of an object is changing over time. It's like knowing if a car is speeding up or slowing down. In calculus, the acceleration function \(a(t)\) is determined by taking the derivative of the velocity function \(v(t)\). If \(v(t)\) describes how position changes over time, \(a(t)\) describes how velocity changes.Consider if the velocity function is \(v(t) = 10t - 3\) for a particular position function. The acceleration is then found by differentiating \(v(t)\), which gives \(a(t) = 10\). This value tells us that the velocity of the object is consistently increasing (or is constant) at this rate.To calculate the acceleration function:
  • Differentiate the velocity function \(v(t)\).
  • This derivative is the acceleration function \(a(t)\).
Understanding acceleration helps predict how objects will behave, especially when force and motion change over time.
Position Function
The position function \(s(t)\) provides a mathematical description of an object's location at any time \(t\). It is crucial in understanding motion along a path. For students, grasping the position function is the foundation for exploring velocity and acceleration.For example, if \(s(t) = 5t^2 - 3t + 15\), it visually indicates how the object moves. The terms in the function describe different characteristics of the motion: - The \(5t^2\) term might suggest accelerated motion.- The \(-3t\) indicates uniform motion.- The constant \(15\) sets an initial position from which movement starts.The position function is always the starting point in any calculus-based motion problem, providing essential insights into how an object is situated over time.
Differentiation Steps
Differentiation is the mathematical process used to find rates of change. In motion problems, it helps us move from determining an object's position to its velocity and then its acceleration. Understanding the differentiation steps is like learning the rules of a game that helps analyze motion at every point along its path.When differentiating a function, the goal is to find another function that represents the rate at which the original function changes. To differentiate a position function like \(s(t)=5t^2-3t+15\), follow these steps:
  • Identify all terms in the function and apply the power rule.
  • Calculate derivatives for individual terms, such as \(5t^2\) becoming \(10t\) and \(-3t\) becoming \(-3\).
  • Combine these derivatives to get the velocity function \(v(t)\).
Repeating the process, differentiate \(v(t)\) to find the acceleration function. Differentiation is foundational in calculus, helping us dissect and predict motion in precise terms.

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Most popular questions from this chapter

A fuel tank is being designed to contain \(200 \mathrm{m}^{3}\) of gasoline, but the maximum length of a tank (measured from the tips of each hemisphere) that can be safely transported to clients is \(16 \mathrm{m}\) long. The design of the tank calls for a cylindrical part in the middle, with hemispheres at each end. If the hemispheres are twice as expensive per unit area as the cylindrical part, find the radius and height of the cylindrical part so the cost of manufacturing the tank will be minimal. Give your answers correct to the nearest centimetre.

Your neighbors operate a successful bake shop. One of their specialties is a very rich whipped-cream-covered cake. They buy the cakes from a supplier who charges \(\$ 6.00\) per cake, and they sell 200 cakes weekly at \(\$ 10.00\) each. Research shows that profit from the cake sales can be increased by increasing the price. Unfortunately, for every increase of \(\$ 0.50\) cents, sales will drop by seven cakes. a. What is the optimal retail price for a cake to obtain a maximum weekly profit? b. The supplier, unhappy with reduced sales, informs the owners that if they purchase fewer than 165 cakes weekly, the cost per cake will increase to \$7.50. Now what is the optimal retail price per cake, and what is the bake shop's total weekly profit? c. Situations like this occur regularly in retail trade. Discuss the implications of reduced sales with increased total profit versus greater sales with smaller profits. For example, a drop in the number of customers could mean fewer sales of associated products.

In a certain manufacturing process, when the level of production is \(x\) units, the cost of production, in dollars, is \(C(x)=3000+9 x+0.05 x^{2}, 1 \leq x \leq 300\) What level of production, \(x,\) will minimize the unit cost, \(U(x)=\frac{C(x)}{x} ?\) Keep in mind that the production level must be an integer.

An object moves so that its velocity, \(v,\) is related to its position, \(s,\) according to \(v=\sqrt{b^{2}+2 g s},\) where \(b\) and \(g\) are constants. Show that the acceleration of the object is constant.

A professional basketball team plays in an arena that holds 20000 spectators. Average attendance at each game has been 14000 . The average ticket price is \$75. Market research shows that for each \(\$ 5\) reduction in the ticket price, attendance increases by \(800 .\) Find the price that will maximize revenue.
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