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At 2:00 p.m. a car's speedometer reads 30 mi/h. At 2:10 p.m. it reads 50 mi/h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mi/h2. v(1/6) – v(0) 1/6 - 0 Let v(t) be the velocity of the car t hours after 2:00 p.m. Then By the Mean Value Theorem, there is a number c such that 0

At 2:00 p.m. a car's speedometer reads 30 mi/h. At 2:10 p.m. it reads 50 mi/h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mi/h2. v(1/6) – v(0) 1/6 - 0 Let v(t) be the velocity of the car t hours after 2:00 p.m. Then By the Mean Value Theorem, there is a number c such that 0

Algebra & Trigonometry with Analytic Geometry
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 p.m. a car's speedometer reads 30 mi/h. At 2:10 p.m. it reads 50 mi/h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mi/h2. Let v(t) be the velocity of the car t hours after 2:00 p.m. Then V(1/6) – v(0) 1/6 - 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 p.m. is exactly 120 mi/h2. Need Help? Read It
Transcribed Image Text:At 2:00 p.m. a car's speedometer reads 30 mi/h. At 2:10 p.m. it reads 50 mi/h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mi/h2. Let v(t) be the velocity of the car t hours after 2:00 p.m. Then V(1/6) – v(0) 1/6 - 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 p.m. is exactly 120 mi/h2. Need Help? Read It
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