Part II: The Answers
320
Answers
401–500
Finding the derivative of the position function gives you the velocity function:
s
t
t
v s
t
=
−
= = −
2
16
2
0
20 32
′
Set the velocity function equal to zero and solve for time t:
20 32 0
20
32
5
8
−
=
=
=
t
t
Substitute in this value to find the height:
s 5
8
20 5
8
16 5
8
25
4
6 25
2
( ) ( ) ( )
=
−
=
= . ft
To find the velocity of the stone when it hits the ground, first determine when the
stone hits the ground by setting the position equation equal to zero and solving for t:
20 16
0
4 5 4
0
0 5
4
2
t
t
t
t
t
−
=
−
=
=
(
)
,
Because t = 0 corresponds to when the stone is first released, substitute t = 5
4
into the
velocity function to get the final velocity:
v = −
= −
20 32 5
4
20
( )
ft/s
496.
downward on 0 2
3
,






 ; upward on 2
3
, ∞






Note that because the velocity is the rate of change in position with respect to time,
you want to find when the velocity is positive and when the velocity is negative.
Assume that when the velocity is positive, the particle is moving upward, and when
the velocity is negative, the particle is moving downward.
Begin by taking the derivative of the position function to find the velocity function:
y t
t
v y
t
= − +
= =
−
3
4 5
′ 3
4
2
Set the velocity function equal to zero and solve for time t:
3
4 0
4
3
2
3
2
t
t
− =
= ±
= ±
Use only the positive solution based on the given interval for t.
Now take a point from the interval 0 2
3
,





 and a point from 2
3
, ∞





 and substitute