Chapter 2

Kinematics: Description of Motion

Important Terms:

Mechanics = the branch of physics concerned with the study of motion and what
produces and affects motion

Kinematics = deals with the description of motion of objects without consideration of

what causes the motion

Dynamics = analyzes the causes of motion

Motion = involves the changing of position

Distance = the total path length traversed in moving from one location to another

Scalar Quantity = one with magnitude only

Speed = the rate at which distance is traveled

Average Speed = the distance traveled divided by the total time elapsed in traveling that
distance

Instantaneous Speed = is how fast something is moving at a particular instant of time

Displacement = the straight-line distance between two points along with the direction
from the starting point to the final position

Vector Quantity = has both magnitude and direction

Velocity = how fast something is moving and in what direction

Average Velocity = displacement divided by the total travel time

Instantaneous Velocity = describes how fast something is moving and in what direction
at a particular instant of time.

Acceleration = time rate of change of velocity

Average Acceleration = the change in velocity divided by the time taken to make the
change

Instantaneous Acceleration = acceleration at a particular instant

Free Fall = objects in motion solely under the influence of gravity
Important Equations:

Average Speed
S = d/t
Where: S = speed
d = distance
t = time

Kinematic Equation for Linear Motion with Constant Acceleration

x = vt
Where:x = displacement
v = velocity
t = time

v=(v f + v i)/2
v f =v i +at
x=v i t +½ a t 2
v f 2=v i2+2 ax

Kinematic Equation Applied to Free Fall

y = vt
v = (vf + vi)/2
vf = vi + gt
y = vit + ½gt2
vf2 = vi2 + 2gy

Example 1: Sojourner could move at a maximum average speed of about 0.6m/min. At

this speed, how long would it take the rover to travel 3m to get to another rock to
analyze?

Given:
m 1min m
s=0.6 × =0.01
min 60 s s
d=3 m

Required:
t=?

Solution:
d
s=
t
m 3m
0.01 =
s t
 3m
t= =300 s
m
0.01
s

Example 2: A jogger jogs from one end to the other of a straight 300m track (from point
A to point B) in 2.5min and then turns around and jogs 100 m back towards the starting
point (to point C) in another 1min. What are the jogger’s average speeds and average
velocities in going from A to B and from B to C after turning around?
Given: 300 m m
S AB= =2
dAB = 300m 150 s s
60 secs
t AB=2.5 mins x =150 s
1 min 100 m m
S BC = =1.67
dBC = 100m 60 s s
60 secs
t BC =1 minx =60 s
1min 300+100 400 m
S AC = = =1.9
150+60 210 s
Required:
SAB = ? 300 m m
V AB= =2
SBC = ? 150 s s
VAB = ?
VBC = ? −100 m m
V BC = =−1.67
60 s s
Solution:

300−100 200 m
V AC = = =0.95
150+60 210 s
Example 3: A couple in their sport utility vehicle is traveling 90km/h down a straight
highway. They saw an accident in the distance, so the driver slows down to 40km/h in
5s. What is the average acceleration of the SUV?

Given:
km 1 hr 1000 m m
V i=90 × × =25
hr 3600 s 1 km s

km 1 hr 1000 m m
V f =40 × × =11.11
hr 3600 s 1 km s

t=5 secs

Required:
a=?

Solution:
 V f −V i 11.11−25 m
a= = =−2.78 2
t 5 s

Example 4: A drag racer starting from rest accelerates in a straight line at a constant rate
of 5.5m/s2 for 6s. What is the racer’s velocity at the end of this time? If a parachute
deployed at this time causes the racer to slow down uniformly at a rate of 2.4m/s 2, how
long will it take the racer to come to a stop?
Given:
m Solution:
V i=0
s a . V f =V i +at =5.5 ×6
m m
a=5.5 2 ¿ 33 =V i
s s
t = 6secs
m b . V f =V i +at
a 2=−2.4 2
s 0=33+ (−2.4 ) t
m m
V f 2=0 33
s s
t=
m
2.4 2
Required: s
a . V f =? t=13.75 s
b . t =?
Example 5: A motorboat starting from rest on a lake accelerates in a straight line at a
constant rate of 3m/s2 for 8s. How far does the boat travel during this time?
Given:
m Required:
V i=0
s x=?
m
a=3 2
s Solution:
t=8 s x=v i t +0.5 a t 2
x=0+ 0.5× 3 ×82

x=96 m
Example 6: A body moves along a straight line following the equation x = 2t 2 – 5t meter
where t is in seconds. Solve the displacement of the body during the time interval from t 3
= 3 sec to t5 = 5s, what is the average velocity of the body, the instantaneous velocities at
this times, average acceleration and instantaneous accelerations?
Given: a. x 3=?
2
x=2 t −5 t b. x 5=?
t 3=3 s c. v=?
t 5=5 s d. v3 =?
e. v5 =?
Required: f. a ave =?
g. a inst=?

Solution:
a . x 3=2 ( 3 )2−5 ( 3 ) b . x 5=2 (5 )2−5 ( 5 )
¿ 18−15 ¿ 50−25
¿3m
∆ d x 5−x 3 25−3 22 m
¿ 25 mc . v= = = = =11
∆ t t 5−t 3 5−3 2 s

dx
=v=4 t 1−5 t 0=4 t−5instantaneous velocity
dt
d . v 3=4 ( 3 )−5
v3 =12−5 e . v 5=4 ( 5 )−5
m v5 =20−5
v3 =7
s
m ∆ v v 5−v 3 15−7 8 m
v5 =15 f . a ave = = = = =4 2
s ∆ t t 5−t 3 5−3 2 s

dv m
g. =a=4 t 0 =4 2 instantaneous acceleration
dt s

Example 7: The motion of a body along a straight line is described by the equation x =
0.2t3 + 3t2 + 2t – 5 meters, where t is in seconds. Solve the displacement of the body
during the time interval from t2 = 2sec to t4 = 4s, what is the average velocity of the body,
the instantaneous velocities at this times, average acceleration and instantaneous
accelerations?
Given: b. x 4 =?
3 2
x=0.2 t +3 t +2t−5 c. v=?
t 2=2 s d. v 2=?
t 4=4 s e. v 4=?
f. a ave =?
Required: g. a 2=?
a. x 2=? h. a 4=?

Solution:
a . x 2=0.2 ( 2 )3+3 ( 2 )2 +2 ( 2 )−5 b . x 4=0.2 ( 4 )3 +3 ( 4 )2+ 2 ( 4 ) −5
¿ 1.6+12+ 4−5 ¿ 12.8+ 48+ 8−5
¿ 12.6 m ¿ 63.8 m

∆ d x 4−x 2 63.8−12.6 51.2 m

c . v= = = = =25.6
∆ t t 4−t 2 4−2 2 s
dx
=v=0.6 ( t )2 +6 t+ 2
dt
d . v 2=0.6 (2)2 +6 ( 2 ) +2
¿ 2.4+12+2 e . v 4=0.6(4 )2+ 6 ( 4 )+2
m ¿ 9.6+ 24+2
¿ 16.4
s
m ∆V v 4−v 2 35.6−16.4 19.2 m
¿ 35.6 f . a ave = = = = =9.6 2
s ∆ t t 4−t 2 4−2 2 s

dv
=a=1.2t +6
dt
g . a2=1.2 ( 2 )+ 6
m h . a4 =1.2 ( 4 )+ 6
¿ 8.4 2
s
m
¿ 10.8 2 Example 8: At the instant the traffic light turns green. An automobile that has
s
been waiting at the intersection starts ahead with constant acceleration of 2m/s 2. At the
same instant a truck, traveling with a constant speed of 18m/s overtakes and passes the
automobile. How far beyond its starting point does the automobile overtake the truck?
How fast is the automobile traveling when it overtakes the truck?
Given:
a C =2 m/s2
V a =0 m/ s
i

a T =0 m/ s2
V T =18 m/s

Required: Xa = XT
X =?

V a =?
f

Solution:

1
x t=V i t + a t 2=V i t=18 t
2
1 1
x a=V i t+ at 2= a t 2=t 2
2 2

x t=x a
18 t=t 2
t=18 s

a . x a=t 2=182=324 m
b . V f =V i +at
m
¿ 2 ( 18 )=36
s

Example 9: What if the car is 20m ahead, the green light turns on.

XT

20m
m Xa

x t=x a +20
18 t=t 2+20
t 2−18 t+20=0
−b ± √❑
t=
❑
t 2=16.81 s
t 1=1.19 s

Example 10:
Time (s) Distance (m)
0 0
1 2
2 3
3 5
4 8
5 12
Velocity(m/s) Time (s) Acceleration (m/s2)
V 1=2 1
V 2=1 2 A1=−1
V 3=2 3 A2=1
V 4 =3 4 A3 =1
V 5=4 5 A 4=1
Example 11:

Time, (sec) Distance, (m)

0 0
1 2
2 4
3 6
4 8
5 10
Velocity, (m/s) Time, (s) Acceleration, (m/s2)
V1 = 2 1
V2 = 2 2 A1 = 0
V3 = 2 3 A2 = 0
V4 = 2 4 A3 = 0
V5 = 2 5
A4=0
Example 12: A ball is dropped out of a window near the top of a building. If the ball
accelerates towards the ground at a rate of 9.81m/s 2, what is its velocity when it has fallen
4.0m?
Given: y=−4 m
m
V i=0
s Required:
m
g=−9.81 2
s
V f =? Solution:
2 2
V f =V i + 2 gy
V f 2=0+2 (−9.81 ) (−4 )
V f =√❑

m
V f =−8.86
s

Example 13: A boy on a bridge throws a stone vertically downward with an initial
velocity of 14.7m/s towards a river below. If the stone hits the water 2s later, what is the
height of the bridge above the water?
Given:
m Required:
V i=−14.7
s
t=2 s
h=?
Solution:
1
y=V i t+ g t 2
2
¿−14.7 ( 2 )+ 0.5 (−9.81 ) (22)
y=−49.02 m
h=49.02 m

Example 14: If a lady grasps the falling ruler as quickly as possible, and the length of the
ruler below the top of the finger is noted. If on the average the ruler descends 18cm
before it is caught, what is the person’s average reaction time?
Given:
y=−0.18 m
m
V i=0
s
g = -9.81m/s2

Required:
t=?

Solution:
1
y=V i t+ g t 2
2
−0.18=0+0.5 (−9.81 ) t 2
t=√ ❑Example 15: A worker on a scaffold on a billboard throws a ball straight up. It has
an initial velocity of 11.2m/s when it leaves his hand at the top of the billboard. What is
the maximum height the ball reaches relative to the top of the billboard? How long does it
take to reach this height? What is the position of the ball at 2s?

Given: ymax

m t=?
V i=11.2
s
V f =0 m/s

Required:
a . y max =?
b . t =?
c . y @2 s=?

Solution:
a . V f 2=V i2+ 2 gy
02 =( 11.2 )2 +2 (−9.81 )( y )
11.2 2
y max = =6.39 m
2 ( 9.81 )

b . V f =V i +¿
0=11.2+ (−9.81 ) t
−11.2
t= =1.14 s
(−9.81 )

c . y =V i t +0.5 g t 2
2
y=11.2 ( 2 ) +0.5 (−9.81 )( 2 )
¿+2.78 m