Solutions to the Unit 2
Review Worksheet
Review for Test

v1  20 ms
v 2  35 ms
v22  v12  2ax


2
2
35  20  2a(100)
a  4.125 sm2
x  12 t(v1  v 2 )
 100  1 t(20  35)
2
t  3.64 sec
x  100m

v1  22 ms
v 2  0 ms
v22  v12  2ax

02 222  2(0.04)x
x  6,050m

v2  v1  at
0  22  (0.04)t
t  550sec
a  0.04 sm2
v1  80 ms
v 2  0 ms
v22  v12  2ax


2
2
0  80  2a(1000)
a  3.2 sm2


x  1000m

v1  0 ms
v 2 11.4 ms
v2  v  2ax

11.4 2  02  2a(16)
2
2
1
x  16m
x  12 t(v1  v 2 )
16  12 t(0 11.4)
t  2.81sec
a  4.06 sm2


x  100m
16m  84m
v1  11.4 ms

x  12 at 2  v1t

84  12 (0)t 2 11.4t
a0
t  7.37sec
total time  2.81sec 
7.37sec  10.18sec
v1  0 ms
v 2  36 ms
x  40m
v22  v12  2ax

362  02  2a(40)
a 16.2 sm2

16.2 sm2
x  t(v1  v 2 )
1
2
40  12 t(0  36)

t  2.22sec

9.8 sm2
 1.65g' s
v1  0
a  9.8 sm2
m
s
t  1.6sec
x  at  v1t


2
1
x  2 (9.8)(1.6)  0(1.6)
1
2
2
x  12.54m

v2  v1  at

v2  0  (9.8)(1.6)
v 2  15.68 ms

h 12.54m

v 3  30
m
s
a  9.8 sm2
x13  30m
2
v3  v1  at
30 17.67  (9.8)t
2
2

  v  2(9.8)(30)
(30)
1
t  4.86sec
v1 17.67 ms

v32  v12  2ax
v1 17.67 ms

a  9.8 sm2
v2  v  2ax
2
2
1
v2  0
m
s
02 17.672  2(9.8)x
haboveground  15.93m 30m  45.93m
1
3
x  15.93m
1-2
1-3
v1  0 ms
a  9.8 sm2
2-3
v1  0 ms
a  9.8 sm2


2
1
x  2 at  v 2 t

 1
18  2 (9.8)(0.76) 2  v 2 (0.76)

m
v 2  19.96 s
1
a  9.8 sm2
x  18m
t  0.76sec
2
3
1-2
1-3
v1  0 ms
a  9.8 sm2


2-3
v1  0 ms
a  9.8 sm2


1
a  9.8 sm2
x  18m
t  0.76sec
v 2  19.96
m
s

2
3
x  12 t(v 2  v 3 ) 18  12 (0.76)(19.96  v 3 )
v 3  27.4 ms
1-2
1-3
v1  0 ms
a  9.8 sm2

2-3
v1  0 ms
a  9.8 sm2
v 3  27.4 ms


2
v32  v
1  2ax
2

2
(27.4)  0  2(9.8)x
x  38.3m
1
a  9.8 sm2
x  18m
t  0.76sec
2
3
h  38.3m
a
v (m/s)
36 ms
v1  36
a0
x  540m

m

9
s2
4 sec
36




4
19
22
x  12 at 2  v1t
540  0  36t
t  15sec
a
36 ms
m

12
s2
3sec
Time (sec)
Area  x  Atriangle  Arectangle Atriangle
 12 (4)(36)  (15)(36)  12 (3)(36)
 666m
3 seconds to stop
v (m/s)
a
3.8 ms
m

19
2
s
0.2sec
3.8

It’s simple physics, really. Taking into
account the amount of drag force
experienced by the car as it leaves the
launching platform, and using a basic
understanding of children’s toys, the color is
most obviously BLACK and the details on
the side doors and hood are YELLOW
FLAMES with RED and ORANGE outlines.
DUH!!!!
?
0.2
v1  3.8 ms
a?
x  12m
v 2  0 ms
Time (sec)
v2  v  2ax
2
2
1
02  3.82  2a(12)
a  0.6 sm2
This is a chase problem!!!
x2A = x2B
Use the chase equation:
1
2
1
2
at 2  v1t  x1  12 at 2  v1t  x1
a(12.2) 2  0(12.2)  0  12 (0)(12.2  5) 2  30(12.2  5)  0

1
2
a(12.2) 2  30(12.2  5)
74.42a  516

a  6.93 sm2
This is another chase problem!!!
x2W = x2T
Use the chase equation:
1
2
at 2  v1t  x1  12 at 2  v1t  x1
1
2
 Graph it!
(0)t 2  8t  0  12 (2.1)t 2  4t  30
8t  1.05t 2  4t  30

0  1.05t 2  4t  30

NO, SHE CANNOT CATCH THE TRAIN.
THE PROBLEM HAS “NO SOLUTION”
(NO REAL ZEROS)
IF she did catch the
train, what would that
look like?
Solve for this
time (maybe 30
seconds, for example)
IF she did catch the train,
how would you find the
distance that she travelled?
1
2
1
2
Plug the time that
you found into
either x2W = x2T
(0)t 2  8t  0  12 (2.1)t 2  4t  30
(0)( 30 ) 2  8( 30 )  0  ?
1
2
( 30 )t 2  4( 30 )  30  ?
Is it possible to have a
problem where there
are TWO (2) times, or
two answers?
What would this kind of a
problem look like?
Solve for both
times
“A man throws a ball straight up
at a speed of 30 ft/s. Find the
times that the ball would be 10
feet above the ground.”
See next page if you
want a solution.
“A man throws a ball straight up at a speed of
30 ft/s. Find the times that the ball would be
10 feet above the ground.”
v1  30
ft
s
a  32.2 sm2
x 10 ft
x  12 at 2  v1t


1
10  2 (32.2)t 2  30t
0  16.1t 2  30t 10


t  0.435sec
t  1.429sec