   Chapter 2.1, Problem 6E

Chapter
Section
Textbook Problem

If a rock is thrown upward on the planet Mars with a velocity of 10 m/ s, its height in meters t seconds later is given by y = 10t – 1.86t2.(a) Find the average velocity over the given time intervals:(i) [1, 2](ii) [1, 1.5](iii) [ 1, 1.1](iv) [ 1, 1.0 1](v) [ 1, 1.001](b) Estimate the instantaneous velocity when t = 1.

(a)

To determine

The average velocity over the given time intervals.

Explanation

Given:

The height (in meters) of the rock after t seconds is y=10t1.86t2.

That is, the position function is y=10t1.86t2.

Calculation:

The average velocity over the time interval [t,t+h] is:

vavg=Change in positionTime elapsed=y(t+h)y(t)(t+h)t=y(t+h)y(t)h

Obtain the average velocity of the position function y=10t1.86t2 over the time interval [t,t+h].

vavg=y(t+h)y(t)h=[10(t+h)1.86(t+h)2][10(t)1.86(t)2]h=[10t+10h1.86(t2+h2+2th)][10t1.86t2]h=10t+10h1.86t21.86h23.72th10t+1.86t2h

Simplify the terms and obtain the vavg as follows.

vavg=10h1.86h23.72thh=h(101.86h3.72t)h=101.86h3.72t

Thus, the average velocity of the position function y=10t1.86t2 over the time interval

[t,t+h] is:

vavg=101.86h3.72t (1)

Section-(i)

Obtain the average velocity over the time interval [1, 2].

Take t=1 and t+h=2 , so that h becomes,

h=2t=21=1

Substitute h=1 and t=1 in equation (1),

vavg=101.86h3.72t=101.86(1)3

(b)

To determine

The instantaneous velocity when t=1.

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