Asked by Zuhayr
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
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
Answers
Answered by
oobleck
0 < c < 1/6
v'(c) is given by the formula at the end of paragraph one.
v'(c) is given by the formula at the end of paragraph one.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.