14. O0/0.26 points | Previous Answers UFLCalc1 1.6.004.Tut. As a certain object

Question

14. O0/0.26 points | Previous Answers UFLCalc1 1.6.004.Tut. As a certain object falls, its position s (in meters) above ground after t seconds is given by s(t) = 25-5t2. (a) what is the average velocity of the object on the interval from t = 1 to the time 0.5 seconds later? X m/s (b) what is the average velocity of the object on the interval from t = 1 to the time 0.1 seconds later? X m/s (c) Write a polynomial in terms of h for the average velocity from t = 1 to the time h seconds later (h # 0). (Simplify your answer completely.) (d) What does this average tend toward for h closer and closer to 0 (smaller and smaller time interval)? X m/s (e) What would you guess to be the exact velocity of the object after 1 second? m/s

Explanation / Answer

(a) Given s(t)=25-5t^2

Average velocity of the object from t=1 to t=0.5 sec later ie t=1.5  

= (s(1.5) - s(1.0))/(1.5 - 1.0)
= (13.75 - 20)/0.5
= -12.5 m/s
(b) Average velocity of the object from t=1 to t=0.1 sec later ie t=1.1
(s(1.1) - s(1.0))/(1.1 - 1.0)
= (18.95 - 20)/0.1
= -10.5 m/s
(c)Average velocity of the object from t=1 to t=h sec later ie t=1+h

(s(1 + h) - s(1))/(1 + h - 1)
= 25 - 5(1+h)² - 20)/h
= {[25 - 5(1 + 2h + h²)] - 20}/h
= (20 - 10h - 5h² - 20)/h
= -5h(2 + h)/h= -10-5h
when h tends to 0

Average velocity= -10m/s
(e) s(t)=25-5t^2

v(t)=ds(t)/dt=-10t

when t=1

v(t)=10m/s

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.