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Answer: -12 ft/sec
f'(x) = [tex]\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]
In this problem, velocity is s'(t): the derivative of the position function.
s'(t) = [tex]\lim_{h \to 0} \dfrac{s(t+h)-s(t)}{h}[/tex]
v(t) = [tex]\lim_{h \to 0} \dfrac{s(t+h)-s(t)}{h}[/tex]
s(t+h) = -16(t+h)² + 20(t+h) + 100
s(t) = -16t² + 20t + 100
v(t) = [tex]\lim_{h \to 0} \dfrac{[-16(t+h)^{2}+20(t+h)+100]-[-16t^{2}+20t+100]}{h}[/tex]
[tex]\lim_{h \to 0} \dfrac{[-16t^{2}-32th-16h^{2}+20t+20h+100]-[-16t^{2}+20t+100]}{h}[/tex]
= [tex]\lim_{h \to 0} \dfrac{-16t^{2}-32th-16h^{2}+20t+20h+100+16t^{2}-20t-100}{h}[/tex]
Several terms cancel out:
= [tex]\lim_{h \to 0} \dfrac{-32th-16h^{2}+20h}{h}[/tex]
Factor out an h from the numerator:
= [tex]\lim_{h \to 0} \dfrac{(-32t-16h+20)h}{h}[/tex]
The h's cancel out:
= [tex]\lim_{h \to 0} \(-32t-16h+20[/tex]
Find the limit. All terms with an h will equal to 0.
= -32t + 20
v(t) = -32t + 20
To find the velocity when t = 1, we need to evaluate v(1):
v(1) = -32(1) + 20
= -32 + 20
= -12 ft/sec