Practice Quiz
1) 15 points A particle moves along a line according to s(t )  2t 3  9t 2  12 where s(t)
represents distance in feet after t seconds. Find:
(a) velocity at t = 2
v(t )  6t 2  18t
v(2)  12 ft / sec
(b) acceleration at t = 1
a(t )  12t  18
v(2)  6 ft / sec 2
(c) when the particle is at rest
6t 2  18t  0
6t (t  3)  0
t  0, t  3
(d) when the particle is moving in a positive direction
after 3 sec
2) 15 points A stone is thrown vertically upward with an initial velocity of 32 ft/sec.
s(t )  16t 2  32t  5 where s is distance in feet from the ground at t seconds.
(a) Find the average velocity on the interval 3 4  t  5 4
s ( 34 )  20
s ( 54 )  20
avg vel 
20  20 0
 1 0
5
3

4
4
2
(b) Find the instantaneous velocity at 3 4 seconds and 5 4 seconds
v(t )  32t  32
v( 3 4 )  8 ft / sec
v( 5 4 )  8 ft / sec
(c) Find the speed at 3 4 seconds and 5 4 seconds
speed : at 3 4 is 8 ft / sec
at 5 4 is 8 ft / sec
(d) How long does it take for the stone to reach its maximum height?
 32
t
 1sec
2(16)
(e) What is the stone’s maximum height?
s(1)  21 ft
(f) When will the stone reach the ground?
 16t 2  32t  5  0
 32  32 2  4(16)(5)
t
2(16)
t  0.15,2.15
hit ground at t  2.15 sec
(g) What is the velocity of the stone when it hits the ground?
v(2.15)  36.8 ft / sec
3) 15 points The following graph shows the velocity of a particle moving along a straight
line.
-1
1
2
3
4
5
6
7
8
9
(a) When does the particle move forward? Move backward? Speed up? Slow
down?
Forward: (0, 1) U (5, 7)
Backward: (1, 5)
Speed Up: (1, 2) U (5, 6)
Slow Down: (0, 1) U (3, 5) U (6, 7)
(b) When is the particle’s acceleration positive? Negative? Zero?
Accel Pos: (3, 6)
Accel Neg: (0, 2) U (6, 7)
Accel Zero: (2, 3) U (7, 9)
(c) When does the particle move with its greatest speed?
t = 0 U (2, 3)
(d) When does the particle stand still for more than an instant?
(7, 9)
Find the second derivative of the function. (5 points each)
4) r ( x)  (2 x  3) 4
r ' ( x)  8(2 x  3) 3
r ' ' ( x)  48(2 x  3) 2