Use the position function s(t)=-16t^{2}+v_{0}t+s_{0} for free-fa

deiteresfp 2021-12-18 Answered
Use the position function s(t)=16t2+v0t+s0 for free-falling objects. A silver dollar is dropped from the top of a building that is 1362 feet tall. (a) Determine the position and velocity functions for the coin. (b) Determine the average velocity on the interval [1, 2]. (c) Find the instantaneous velocities when t=1 and t=2. (d) Find the time required for the coin to reach ground level. (e) Find the velocity of the coin at impact.
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Expert Answer

godsrvnt0706
Answered 2021-12-19 Author has 31 answers

Step 1
a) The position function:
in order to determine this equation we have to determine v0 and s0:
the coin is dropped with no initial velocity (free falling object) so we conclude that v0=0.
When t=0 have s(0)=s0=1362 (because the coin was dropped from the top of the building with a height =1362)
s(t)=16t2+v0t+s0
v0 is the initial velocity (when t=0)
s0 is the initial position (when t=0)
(these concepts are from physics!)
we conclude that: the position function =s(t)=16t2+1362
Step 2
a) the velocity function =d(s(t))dt
v(t)=d(s(t))dt=32t
Step 3
b) the average velocity =s(2)s(1)21=s(2)s(1)
the average velocity =s(2)s(1)=16×4+1362(16×1+1362)=48
Step 4
c) the instantaneous velocity at 2 is v(2)
the instantaneous velocity at 1 is v(1)
v(2)=32×2=64
v(1)=32
Step 5
d) time required to the coin to reach the ground:
when coin reach the ground we have s(t)=0
so we must solve this equation to determine the time
s(t)=0
16t2+1362=0
16t2=1362
t2=136216=85.125
t=85.1259.23
Step 6
e) plug the time when the coin reaches the ground into the velocity equation: v(t)=32t
v(85.125)v(9.23)=32×9.23=295.36
Answer:
a) s(t)=16t2+1362,v(t)=32t
b) the average velocity =48
c) v(1)=32,v(2)=64
d) t9.23
e) t295.36

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Elaine Verrett
Answered 2021-12-20 Author has 41 answers

Good grief, feet and inches and yards. That is an old textbook with
g=32ftsecond2 so g2=16
anyway s=(12)ft2+Vot+Ho
so a) s=16t2+0+1362
dsdt=velocity=32t+0
and of course acceleration =d2sdt2=32
b) find out how far it moved in that one second and divide by one second
s(2)=164+1362
s(1)=161+1362
difference =48 ft in one second so average speed =48 ft/second beteen t=1 and t=2
c) v(1)=32(1)=32fts
v(2)=32(2)=64fts
d) when will s=0?
0=16t2+0+1362
16t2=1362
t2=681
t=26.1 seconds
e) 16(26.1)=32(26.1)

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