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Q34E

Expert-verifiedFound in: Page 216

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**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.

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

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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