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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/h2. Let v(t) be the velocity of the car t hours after 2:00 PM. 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 PM is exactly 120 mi/h2.

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rodolforodriguezr

Answer:

See proof below

Step-by-step explanation:

If we consider 2 PM as hour 0, then 10 minutes later would be hour 1/6 (since 10 minutes equals 1/6 hours).

If V(t) is the speed at hour t, because of the Mean Value Theorem, there exists a point c in [0, 1/6] such that

[tex]\bf V'(c)=\frac{V(1/6)-V(0)}{1/6-0}=6(V(1/6)-V(0))[/tex]

But

V(0) = 30 m/h and V(1/6) = 50 m/h, hence

[tex]\bf V'(c)=6(V(1/6)-V(0))=6(50-30)=6*20=120[/tex]

Since V'(c) is the acceleration at hour c, the proof is complete.

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