Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

At 2:00 PM a car’s speedometer reads \(30 mi/h\). At 2:10 PM it reads \(50 mi/h\). Show that at some time between 2:00 and 2:10 the acceleration is exactly \(120 mi/h\)

The car achieves the acceleration of \(120{\rm{mi}}/{{\rm{h}}^2}\) at some time between \(2:00\) and \(2:10\)is proved.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

"If \(f\) be a function that satisfies the following hypothesis:

1. \(f\) is continuous on the closed interval \((a,\;b)\).

2. \(f\) is differentiable on the open interval \((a,\;b)\).

Then, there is a number \(c\) in \((a,\;b)\) such that \({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently, \(f(b) - f(a) = {f^\prime }(c)(b - a)\)”.

Step 3: Prove that at some time between \(2:\;00\) and \(2:\;10\) the acceleration of the car is exactly \(120{\rm{mi}}/{{\rm{h}}^2}\)

The velocity of the car \(v\) is continuous on the time interval \((2:00{\rm{pm}},\;2:10{\rm{pm}})\) and derivative of the velocity is the acceleration.

Acceleration of the car exists always, even if the car is stopped, the acceleration must be only \(0\). So, always it exists.

Thus, the velocity \(v\) is differentiable on the interval \((2:00{\rm{pm}},\;2:10{\rm{pm}})\).

Use the Mean Value Theorem here, by that there must be some time \(t\) in the time interval \((2:00{\rm{pm}},\;2:10{\rm{pm}})\) such that \({v^\prime }(t) = \frac{{v(2:10) - v(2:00)}}{{2:10 - 2:00}}\).

Substitute the corresponding values of the velocity in the above equation as follows:\({v^\prime }(t) = \frac{{50{\rm{mi}}/{\rm{h}} - 30{\rm{mi}}/{\rm{h}}}}{{2:10 - 2:00}}\)

Since derivative of the velocity is acceleration, it is denoted by \(a(t)\) as follows:

\(\begin{aligned}{c}a(t) &= \frac{{50{\rm{mi}}/{\rm{h}} - 30{\rm{mi}}/{\rm{h}}}}{{10{\rm{mi}}}}\\ &= \frac{{20{\rm{mi}}/{\rm{h}}}}{{10{\rm{mi}}}}\\ &= \frac{{20{\rm{mi}}/{\rm{h}}}}{{\frac{1}{6}h}}\\ &= 120{\rm{mi}}/{{\rm{h}}^2}\end{aligned}\)

Hence, it is showed that the car achieves the acceleration of \(120{\rm{mi}}/{{\rm{h}}^2}\) at some time between \(2:00\) and \(2:10\).

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

At 2:00 PM a car’s speedometer reads \(30 mi/h\). At 2:10 PM it reads \(50 mi/h\). Show that at some time between 2:00 and 2:10 the acceleration is exactly \(120 mi/h\)

Step by Step Solution

Step 1: Given data

Step 3: Prove that at some time between \(2:\;00\) and \(2:\;10\) the acceleration of the car is exactly \(120{\rm{mi}}/{{\rm{h}}^2}\)

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

At 2:00 PM a car’s speedometer reads \(30 mi/h\). At 2:10 PM it reads \(50 mi/h\). Show that at some time between 2:00 and 2:10 the acceleration is exactly \(120 mi/h\)

Step by Step Solution

Step 1: Given data

Step 3: Prove that at some time between \(2:\;00\) and \(2:\;10\) the acceleration of the car is exactly \(120{\rm{mi}}/{{\rm{h}}^2}\)

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