Question

suppose that as an object falls from the top of a cliff, its position in feet above the ground after ( t ) seconds is given by ( s(t) = 160 - 16t^2 ). find the average velocity from ( t = 1 ) to ( t = 1 + h ) seconds, where ( h
eq 0 ) ( circ -32 - 16h ) ( circ -32 + 16h ) ( circ 32 - 16h ) ( circ 32 + 16h ) ( circ \frac{-16h^2 - 32h - 32}{h} )

Explanation:

Step1: Recall average velocity formula

The average velocity \( v_{avg} \) over the interval \([a, b]\) is given by \( v_{avg}=\frac{s(b)-s(a)}{b - a} \). Here, \( a = 1 \) and \( b=1 + h \), so we need to find \( s(1 + h) \) and \( s(1) \) first.

Step2: Calculate \( s(1) \)

Substitute \( t = 1 \) into \( s(t)=160-16t^{2} \):
\( s(1)=160-16\times(1)^{2}=160 - 16=144 \)

Step3: Calculate \( s(1 + h) \)

Substitute \( t = 1+h \) into \( s(t)=160-16t^{2} \):
\( s(1 + h)=160-16\times(1 + h)^{2} \)
Expand \( (1 + h)^{2}=1 + 2h+h^{2} \):
\( s(1 + h)=160-16\times(1 + 2h+h^{2})=160-16-32h - 16h^{2}=144-32h - 16h^{2} \)

Step4: Compute \( s(1 + h)-s(1) \)

Subtract \( s(1) \) from \( s(1 + h) \):
\( s(1 + h)-s(1)=(144-32h - 16h^{2})-144=-32h - 16h^{2} \)

Step5: Compute the average velocity

The time interval is \( (1 + h)-1=h \), so average velocity \( v_{avg}=\frac{s(1 + h)-s(1)}{(1 + h)-1}=\frac{-32h - 16h^{2}}{h} \)
Factor out \( h \) from the numerator (since \( h
eq0 \), we can divide by \( h \)):
\( v_{avg}=\frac{h(-32 - 16h)}{h}=-32 - 16h \)

Answer:

\(-32 - 16h\) (corresponding to the first option)