Question

An object is thrown directly upwards. Its height is h(t) = -16t^2 + 64t feet after t seconds. What is the average velocity from t = 1 to t = 2? What is the average velocity from t = 1 to t = 1 + h? Use the result in (b) and a limiting process to find the instantaneous velocity at t = 1.

Name: an object is thrown directly upwards its height is ht 16t264t feet after t seconds what is the average velocity from t1 to t2 what is the average velocity from t1 to t1h use the result in b 23797
Uploaded: 2023-03-01T02:39:42-08:00
Duration: 3 min 32 s
Channel: Gregory Higby
Description: an object is thrown directly upwards its height is ht 16t264t feet after t seconds what is the average velocity from t1 to t2 what is the average velocity from t1 to t1h use the result in b 23797

          An object is thrown directly upwards. Its height is
h(t) = -16t^2 + 64t feet after t seconds.
What is the average velocity from t = 1 to t = 2?
What is the average velocity from t = 1 to t = 1 + h?
Use the result in (b) and a limiting process to find the
instantaneous velocity at t = 1.

Added by Kelly A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Gregory Higby

03/01/2023

Step 1

The average velocity is given by the change in height divided by the change in time. So, we have: Average velocity from t=1 to t=2 = $\frac{h(2)-h(1)}{2-1}$ Now, we need to find the height at t=1 and t=2: $h(1) = -16(1)^2 + 64(1) = -16 + 64 = 48$ $h(2) = Show more…

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An object is thrown directly upwards. Its height is h(t) = -16t^2 + 64t feet after t seconds. What is the average velocity from t = 1 to t = 2? What is the average velocity from t = 1 to t = 1 + h? Use the result in (b) and a limiting process to find the instantaneous velocity at t = 1.