Tutorial Exercise Use the position function s(t) = -4.9t2 + 750, which gives the height (in meters) of an object that has fallen for t seconds from a height of 750 meters. The velocity at time t = a seconds is given by the fol s(a)-s(t) lim Find the velocity of the object when t = 2. Step 1 The velocity of the object when t = a = 2)-s(t) lim 2-t =730.4 Step 2 Find s(2) by substituting 2 for t in s(t) = -4.9t2 + 750. s(2) = -4.9(22)² + 750 730.4 Step 3 Find the value of s(2)-s(t). 2 2 Is s(2)-s(t) = -19.6 + 750-(-4.9t² + 750) Submit Skip (you cannot come back)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.6: Solving Quadratic Equations By Using The Quadratic Formula
Problem 64PFA
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Tutorial Exercise
Use the position function s(t) = -4.9t² + 750, which gives the height (in meters) of an object that has fallen for t seconds from a height of 750 meters. The velocity at time t = a seconds is given by the following.
s(a) s(t)
a-t
Find the velocity of the object when t = 2.
Step 1
The velocity of the object when t = a = 2 2 Is
s(2
2
) -s(t)
Step 2
Find s(2) by substituting 2 fort in s(t) = -4.9t2 + 750.
s(2) = -4.9(2
2)² + 750
= 730.4
730.4
Step 3
Find the value of s(2) - s(t).
s(2) s(t)=19.6+750 (-4.9t² + 750)
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Transcribed Image Text:Tutorial Exercise Use the position function s(t) = -4.9t² + 750, which gives the height (in meters) of an object that has fallen for t seconds from a height of 750 meters. The velocity at time t = a seconds is given by the following. s(a) s(t) a-t Find the velocity of the object when t = 2. Step 1 The velocity of the object when t = a = 2 2 Is s(2 2 ) -s(t) Step 2 Find s(2) by substituting 2 fort in s(t) = -4.9t2 + 750. s(2) = -4.9(2 2)² + 750 = 730.4 730.4 Step 3 Find the value of s(2) - s(t). s(2) s(t)=19.6+750 (-4.9t² + 750) Skip (you cannot come back)
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