EXAMPLES/NOTES:
UNIFORM RECTILINEAR MOTION
1. Car A at a gasoline station stays there for 10 minutes after Car B passes at a constant speed of 40 miles
per hour. How long will it take Car A to overtake Car B if it accelerates at 4 m/s²
Solution:
Ta = Tb
Sa = S10 + Sb
Consider B:
v
=
s
t
Tb
=
s
v
=
Sb
17.88
m
/
s
Sb
=
17.88
Tb
Consider A:
Sa
=
VoTa
+
1
2
aT
²
=
0
+
1
2
(
4
m
s
2
)
(
T a
2
)
=
2
Ta
²
For S10:
S
10
=
vt
=
(
17.88
m
s
)
(
600
s
)
=
10728
m
Substitute:
Sa
=
S
10
+
Sb
2
T
²
=
10728
+
17.88
T

Answer: T = 77.85 seconds
2. A stone is thrown up from the ground with a velocity of 300ft/s. How long must one wait before
dropping a second stone from the top of 600ft tower if the two stones are to pass each other 200ft from
the top of the tower?
Solution:
Consider
1
s
=
Vot
−
1
2
>
²
400
ft
=
(
300
ft
s
)
(
t
)
−
1
2
(
32.2
ft
s
2
)
(
t
2
)
t
=
17.19
s ,
1.45
s
Consider 2
s
=
Vot
+
1
2
gt
2
200
=
0
+
1
2
(
32.2
ft
s
2
)
(
t
2
)
t
=
±
3.52
s
How long must one wait?
t = 17.19 seconds – 3.52 seconds
Answer: 13.67 seconds
VARIABLE ACCELERATION
1. The motion of a particle is given by
s
=
2
t
4
−
t
3
6
+
2
t
2
where
s
is in
feet
and
t
in
seconds
. Compute the values of
v
and
a
when
t
=
2
seconds
.

s
=
2
t
4
−
t
3
6
+
2
t
2
v
=
8
t
3
−
1
2
t
2
+
4
t
a
=
24
t
2
−
t
+
4
@t
=
2
seconds
v
=
8
(
2
)
3
−
1
2
(
2
)
2
+
4
(
2
)
;
v
=
70
ft
sec
a
=
24
(
2
)
2
−
(
2
)
+
4
;
a
=
98
ft
sec
2
2. A particle moves in a straight line according to the law
s
=
t
3
−
40
t
where
s
is in
m
and
t
in
seconds
.
a.
When
t
=
5
seconds
, compute
v
.
b.
Find the average velocity during the 3
rd
to 4
th
seconds.
c.
When the particle comes to stop, what is its acceleration?
a.
s
=
t
3
−
40
t
v
=
3
t
2
−
40
a
=
6
t
@t
=
5
seconds
v
=
3
(
5
)
2
−
40
;
v
=
35
m
sec

b.
v
ave
=
total distance
totaltime
=
−
96
−(−
93
)
4
−
3
=
−
3
1
;
v
ave
=−
3
m
sec
@
3
rd
second
s
=
(
3
)
3
−
40
(
3
)
;
s
=−
93
m
@
4
th
second
s
=(
4
)
3
−
40
(
4
)
;
s
=−
96
m
c.
if v
=
0
0
=
3
t
2
−
40
t
=
3.65
seconds
a
=
6
t
=
6
(
3.65
)
;
a
=
21.9
m
sec
2
3. The rectilinear motion of a given particle is given by
s
=
v
2
−
9
where
s
is in
m
and
v
is
in
m
sec
. When
t
=
0
,
s
=
0
, and
v
=
3
m
sec
. Determine
s
−
t
,
v
−
t
, and
a
−
t
relations.
s
=
v
2
−
9
ds
dt
=
2
v
dv
dt
v
=
2
va
a
=
1
2
a
=
v
−
v
o
t

1
2
=
v
−(
3
)
t
;
v
=
1
2
t
+
3
s
=
v
o
t
+
1
2
at
2
=
(
3
)
t
+
1
2
(
1
2
)
t
2
;
s
=
3
t
+
1
4
t
2
MOTION CURVES
1. A particle starting with an initial velocity of 60 ft /s has a rectilinear motion with the constant
deceleration of 10 ft/
s
2
. Determine the velocity and displacement at t = 9 sec.
Solution:

For
and
:
2. An auto travelled 1800 ft in 40sec. The auto accelerates uniformly and decelerates uniformly at
6 ft/
s
2
, starting from rest at A and coming to stop at B. Find the maximum speed in fps.
Solution:

120
)
V
A
6x – 6(40 – x) = 0
6x – 0 = 6(40 – x)
6x = 240 – 6x
x = 20
3. An Auto starts from rest and reaches a speed of 60 ft/s in 15 sec. The acceleration increases uniformly
from zero for the first 9 sec after which the acceleration reduces uniformly to zero in the next 6 sec.
Compute for the displacement in this 15 sec interval.
Solution:

V:
For
and
:

PROJECTILE MOTION
1. A golf ball is fired from the top of a cliff 50 m high with a velocity of 10 m/s directed at 45⁰ to the
horizontal. Find the range of the projectile.