Introduction to linear kinematics

1. An
Introduction
to Linear
Kinematics
LINEAR KINEMATICS

2. Kinematic Analysis
Linear Kinematics
description of the motion of a body
the appearance of a motion with respect to time
Motion described in terms of (variables):
Distance, displacement, length (e.g. stride, stroke)
Time, cadence (e.g. stride frequency, stroke frequency)
Speed, velocity
Acceleration
Single point models
e.g. Centre of mass (CM) during running/jumping
Multi-segment models
e.g. Co-ordination of body segments during running/jumping

3. Distance & Displacement
Distance:
Length of path which a body covers during motion
Units: metre (m), centimeter (cm), kilometer (km)
Displacement:
The change in position of a body during motion
Units: metre (m), centimeter (cm), kilometer (km)
Distance is a scalar, and displacement is a vector variable

4. Speed and Velocity
Speed (scalar)
Length of path (distance)
divided by change in time
(∆t)
Average velocity (vector)
Change in position (∆p)
divided by change in time
(∆t)
Displacement (d) divided
by change in time (∆t)
Vector equivalent of linear
speed
If displacement = 50 m
Δt
d
=
Δt
Δp
=v
If Dt = 5 s
v = 50 / 5
= 10 m·s
-1

5. Velocity
Units of velocity
m/s or m·s
-1
Velocity is a vector
Magnitude and direction
calculated using
Pythagoras and
trigonometry
The velocity of a swimmer
in a river is the vector sum
of the velocities of
swimmer and current.
Current
velocity
Swimmer’s
velocity
Resultant
velocity

6. 6
1.A speedboat moving at 30.0 m s
-1
approaches stationary
buoy marker 100 m ahead. The pilot slows the boat with a
constant acceleration of -3.50 m s
-2
by reducing the throttle.
a. How long does it take the boat to reach the buoy?
b. What is the velocity of the boat when it reaches the buoy?
No. 23,pg. 51,Physics for scientists and engineers with
modern physics, Serway & Jewett,6
th
edition.
ANS. : 4.53 s; 14.1 m sANS. : 4.53 s; 14.1 m s
--11
2.An unmarked police car travelling a constant 95 km h
-1
is
passed by a speeder traveling 140 km h
-1
. Precisely 1.00 s
after the speeder passes, the policemen steps on the
accelerator; if the police car’s acceleration is 2.00 m s
-2
, how
much time passes before the police car overtakes the
speeder (assumed moving at constant speed)?
No. 44, pg. 41,Physics for scientists and engineers with
modern physics, Douglas C. Giancoli,3
rd
edition.
ANS. : 14.4 sANS. : 14.4 s
Exercise 2.2 :

7. 7
A ball is thrown from the top of a building is given an initial velocity
of 10.0 m s
-1
straight upward. The building is 30.0 m high and the
ball just misses the edge of the roof on its way down, as shown in
figure 2.7. Calculate
a. the maximum height of the stone from point A.
b. the time taken from point A to C.
c. the time taken from point A to D.
d. the velocity of the ball when it reaches point D.
(Given g = 9.81 m s
-2
)
Example 2.7 :
A
B
C
D
u =10.0 m s
-1
30.0 m
Figure 2.7Figure 2.7

8. 2.4. Projectile motion
A projectile motion consists of two components:
vertical component (y-comp.)
motion under constant acceleration, a
y
= -g
horizontal component (x-comp.)
motion with constant velocity thus a
x
= 0
The path followed by a projectile is called trajectory is
shown in Figure 2.9.
8
v
u
q
s
x
= R
s
y
=H
u
x
v
2y
u
y
v
1x
v
1y
v
2x
v
1
q
1
v
2
q
2
t
1
t
2
B
A
P
Q
C
y
x
Figure 2.9Figure 2.9
Simulation 2.5

9. 9
From the trigonometry identity,
thus

The value of R maximum when qq = = 4545°° and sin 2sin 2qq = = 11
therefore
qqq cossin22sin=
q2sin
2
g
u
R=
g
u
R
2
max
=
Simulation 2.6

10. 10
Figure 2.10 shows a ball bearing rolling off the end of a table
with an initial velocity, u in the horizontal direction.
Horizontal component along path AB.
Vertical component along path AB.
2.4.5 Horizontal projectile
h
x
A B
u u
v
x
v
yv
Figure 2.10Figure 2.10
constant velocity, ===
xx vuu
xs
x= nt,displaceme
0u
y
= velocity,initial
hs
y-= nt,displaceme
Simulation 2.7

11. 11
Figure 2.12 shows a ball thrown by superman
with an initial speed, u = 200 m s
-1
and makes an
angle, q = 60.0° to the horizontal. Determine
a. the position of the ball, and the magnitude and

direction of its velocity, when t = 2.0 s.
Example 2.9 :
Figure 2.12Figure 2.12 x
O
u
q = 60.0°
y
R
H
v
2y
v
1x
v
1y v
2x
Q
v
1
P
v
2

12. 12
b. the time taken for the ball reaches the maximum height, H and
calculate the value of H.
c.the horizontal range, R
d. the magnitude and direction of its velocity when the ball
reaches the ground (point P).
e. the position of the ball, and the magnitude and direction of its
velocity at point Q if the ball was hit from a flat-topped hill with
the time at point Q is 45.0 s.
(Given g = 9.81 m s
-2
)
Solution :Solution :
The component of Initial velocity :
1
s m 1000.60cos200
-
==

x
u
1
s m 1730.60sin200
-
==

yu

13. 13
Figure 2.13Figure 2.13
Use gravitational acceleration, g = 9.81 m s
-2
1.A basketball player who is 2.00 m tall is standing on the floor
10.0 m from the basket, as in Figure 2.13. If he shoots the
ball at a 40.0° angle above the horizontal, at what initial
speed must he throw so that it goes through the hoop without
striking the backboard? The basket height is 3.05 m.
ANS. : 10.7 m sANS. : 10.7 m s
--11
Exercise 2.4 :

14. 14
2.An apple is thrown at an angle of 30° above the horizontal
from the top of a building 20 m high. Its initial speed is
40 m s
-1
. Calculate
a. the time taken for the apple to strikes the ground,
b. the distance from the foot of the building will it strikes
the ground,
c.the maximum height reached by the apple from the
ground.
ANS. : 4.90 s; 170 m; 40.4 mANS. : 4.90 s; 170 m; 40.4 m
3.A stone is thrown from the top of one building toward a tall
building 50 m away. The initial velocity of the ball is 20 m s
-1

at 40° above the horizontal. How far above or below its
original level will the stone strike the opposite wall?
ANS. : 10.3 m below the original level.ANS. : 10.3 m below the original level.
Exercise 2.4 :

15. Acceleration
Acceleration = change in Velocity / time
(How fast you change how fast your going!)
Vector = Magnitude (size) + Direction
Units: mi/hr/sec; m/s/s; m/s
2
f iv vv
a
t t
-D
= =
v
v
a = accelerationa = acceleration
DDv = change in velocity (final – initial)v = change in velocity (final – initial)
t = timet = time

16. Constant acceleration woman covers more
distance every second in the same direction
her change in speed is what’s constant!

17. Free Fall Speeds &
Distances
How FAST = Rearrange acceleration equation to solve for v
f
:
f iv v
a
t
-
=
v
f iv v
at
-
=
v
f
i
v
at v+ =
v
f iv v at= +
v
How How FARFAR = equation for free fall distances when falling from rest: = equation for free fall distances when falling from rest:
21
2
x atD =
v
DDx = change in position (displacement) x = change in position (displacement)
a = accelerationa = acceleration
t = timet = time

18. Summary
Variables used to describe motion are either:
Scalar (magnitude only: e.g. time,
distance and speed)
Vector (magnitude and direction: e.g.
displacement, velocity and acceleration)
Displacement is the change in position of a
body
Average velocity is the change in position
divided by the change in time
Average acceleration is the change in
velocity divided by the change in time

19. 19
THE END…
Linear motion

20. Recommended Reading
Enoka, R.M. (2002). Neuromechanics of
Human Movement (3rd edition).
Champaign, IL.: Human Kinetics. Pages 3-
10 & 22-27.
Hamill, J. & Knutzen, K.M. (2003).
Biomechanical Basis of Human
Movement (2nd edition). Philadelphia:
Lippincott Williams & Wilkins. Pages 271-
289.
www.dboccio.com/Physics
www.learnconceptualphysics.com/resources/po
werpoint/ch2-linear_motion
www.ux1.eiu.edu/