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Expert-verified Found in: Page 216 ### Essential Calculus: Early Transcendentals

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.

See the 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 $$2:10$$. ### Want to see more solutions like these? 