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Essential Calculus: Early Transcendentals
Found in: Page 216
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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.

See the step by step solution

Step by Step Solution

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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