At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acceleration is exactly 80 mi/h2. Let v(t) be the velocity of the car t hours after 2:00 PM. Then V(1/4) – v(0) 1/4 – 0 By the Mean Value Theorem, there is a number c such that 0

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.7: More On Inequalities
Problem 44E
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At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and
2:15 the acceleration is exactly 80 mi/h2.
Let v(t) be the velocity of the car t hours after 2:00 PM. Then V(1/4) – v(0)
1/4 – 0
By the Mean
Value Theorem, there is a number c such that 0 <c <
with v'(c) =
Since
v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 80 mi/h2.
Transcribed Image Text:At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acceleration is exactly 80 mi/h2. Let v(t) be the velocity of the car t hours after 2:00 PM. Then V(1/4) – v(0) 1/4 – 0 By the Mean Value Theorem, there is a number c such that 0 <c < with v'(c) = Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 80 mi/h2.
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