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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 2, Problem 2.1.17

Consider the position function s(t)=−16t^2+128t (Exercise 13). Complete the following table with the appropriate average velocities. Then make a conjecture about the value of the instantaneous velocity at t=1. <IMAGE>

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Step 1: Understand the problem. The position function s(t) = -16t^2 + 128t represents the position of an object at time t. We need to find average velocities over certain intervals and make a conjecture about the instantaneous velocity at t = 1.
Step 2: Recall the formula for average velocity over an interval [a, b], which is given by (s(b) - s(a)) / (b - a).
Step 3: Calculate the average velocity over the interval [1, 1+h] for small values of h. This involves computing (s(1+h) - s(1)) / h.
Step 4: Substitute s(t) = -16t^2 + 128t into the average velocity formula: (s(1+h) - s(1)) / h = ((-16(1+h)^2 + 128(1+h)) - (-16(1)^2 + 128(1))) / h.
Step 5: Simplify the expression from Step 4 and evaluate the limit as h approaches 0 to conjecture the instantaneous velocity at t = 1.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Position Function

A position function describes the location of an object at a given time, typically represented as s(t). In this case, s(t) = -16t^2 + 128t models the vertical position of an object under the influence of gravity, where t is time in seconds. Understanding this function is crucial for analyzing motion and calculating velocities.
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Relations and Functions

Average Velocity

Average velocity is defined as the change in position over the change in time, calculated as (s(t2) - s(t1)) / (t2 - t1). It provides a measure of how fast an object is moving over a specific interval. In the context of the given position function, calculating average velocities at different intervals helps in understanding the object's overall motion.
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Derivatives Applied To Velocity

Instantaneous Velocity

Instantaneous velocity is the velocity of an object at a specific moment in time, represented mathematically as the derivative of the position function, v(t) = s'(t). It provides a precise measure of how fast the object is moving at that exact time. Making a conjecture about the instantaneous velocity at t=1 involves evaluating the derivative of the position function at that point.
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Related Practice
Textbook Question

Use a graph of f to estimate limxaf(x){\(\displaystyle\]\lim\)_{x\(\to\) a}}f\(\left\)(x\(\right\)) or to show that the limit does not exist. Evaluate f(x) near x=ax=a to support your conjecture.

f(x)=1cos(2x2)(x1)2;a=1f\(\left\)(x\(\right\))=\(\frac{1-\cos\left(2x-2\right)}{\left(x-1\right)^2}\);a=1

Textbook Question

Determine limxf(x)\(\lim\)_{x\(\rightarrow\]\infty\)}f\(\left\)(x\(\right\)) and limxf(x)\(\lim\)_{x\(\rightarrow\)-\(\infty\)}f\(\left\)(x\(\right\)) for the following functions. Then give the horizontal asymptotes of ff (if any).


f(x)=40x5+x216x42xf\(\left\)(x\(\right\))=\(\frac{40x^5+x^2}{16x^4-2x}\)

Textbook Question

The following table gives the position s(t)s\(\left\)(t\(\right\)) of an object moving along a line at time tt. Determine the average velocities over the time intervals [1,1.01]\(\left\[\lbrack\)1,1.01\(\right\]\rbrack\), [1,1.001]\(\left\[\lbrack\)1,1.001\(\right\]\rbrack\), and [1,1.0001]\(\left\]\lbrack\)1,1.0001^{}\(\right\).]. Then make a conjecture about the value of the instantaneous velocity at t=1t=1. <IMAGE>

Textbook Question

Evaluate each limit and justify your answer. 

lim x→1 (x+5x / x+2)^4

Textbook Question

Determine the points on the interval (0, 5) at which the following functions f have discontinuities. At each point of discontinuity, state the conditions in the continuity checklist that are violated. <IMAGE>

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Textbook Question

Determine limxf(x)\(\lim\)_{x\(\rightarrow\]\infty\)}f\(\left\)(x\(\right\)) and limxf(x)\(\lim\)_{x\(\rightarrow\)-\(\infty\)}f\(\left\)(x\(\right\)) for the following functions. Then give the horizontal asymptotes of ff (if any).


f(x)=4x3+12x3+16x6+1f\(\left\)(x\(\right\))=\(\frac{4x^3+1}{2x^3+\sqrt{16x^6+1}\)}

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