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=1to t=1+h seconds, where h not= 0.

a.32+16h

b.-32-16h

c.32-16h

d.-32+16h

Question

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Respuesta :

cryssatemp cryssatemp · Answer 1 · 2019-07-04T22:48:02+02:00

Explanation:

Velocity [tex]V[/tex] is defined as the variation of the position [tex]s[/tex] of an object in time [tex]t[/tex]:

[tex]V=\frac{\Delta s}{\Delta t}[/tex] (1)

In this case we are given the position above the ground of an object after t seconds by:

[tex]s_{(t)}=160-16t^2[/tex] (2)

When [tex]t=1[/tex]:

[tex]s_{(1)}=160-16(1)^2=144[/tex] (3)

When [tex]t=1+h[/tex]:

[tex]s_{(1+h)}=160-16(1+h)^2=160-16(1+2h+h^2)[/tex] (4)

[tex]s_{(1+h)}=160-16-32h-16h^2=144-32h-16h^2[/tex] (5)

Calculating [tex]\Delta s[/tex]:

[tex]\Delta s=s_{(1+h)}-s_{(h)}=144-32h-16h^2-144[/tex] (6)

[tex]\Delta s=h(-32-16h)[/tex] (7)

Calculating [tex]\Delta t[/tex]:

[tex]\Delta t=1+h-1=h[/tex] (8)

Substituting (7) and (8) in (1):

[tex]V=\frac{h(-32-16h)}{h}[/tex] (9)

Finally:

[tex]V=-32-16h[/tex]