analysing of motion1
TRANSCRIPT
-
8/6/2019 Analysing of Motion1
1/57
Analysing of Motion
-
8/6/2019 Analysing of Motion1
2/57
Learning Outcomes
Determine displacement, average velocity
and acceleration.
-
8/6/2019 Analysing of Motion1
3/57
A ticker timer is a device used to study
motion.
It can print dots on a tape at a steady rate.
The distance between dots on a ticker tape
represents the object's position change
during that time interval.
-
8/6/2019 Analysing of Motion1
4/57
A large distance between dots indicates that
the object was moving fast during that time
interval.
A small distance between dots means the
object was moving slow during that timeinterval.
Ticker tapes for a fast-moving and slow-moving object are depicted below.
-
8/6/2019 Analysing of Motion1
5/57
-
8/6/2019 Analysing of Motion1
6/57
1.
The distance between two neighbouring
dots are the same throughtout the tape.
Therefore the trolley moved with uniform
velocity.
-
8/6/2019 Analysing of Motion1
7/57
2.
The distances between two neighbouring dotsare increasing.
Therefore the trolley moved with increasing
velocity.
The trolley was accelerating.
-
8/6/2019 Analysing of Motion1
8/57
3.
The distances between two neighbouring
dots are decreasing.
Therefore, the trolley moved with decreasing
velocity.
The trolley was decelerating.
-
8/6/2019 Analysing of Motion1
9/57
-
8/6/2019 Analysing of Motion1
10/57
-
8/6/2019 Analysing of Motion1
11/57
-
8/6/2019 Analysing of Motion1
12/57
-
8/6/2019 Analysing of Motion1
13/57
Determination of displacement as the length of
ticker tape over a period of time.
xy = displacement over time t
t = 7 ticks
= 0.14 s
x y
. . . . . . . .
-
8/6/2019 Analysing of Motion1
14/57
Determination of Velocity
. . . . . . . .
12 cm
Displacement = 12 cmTime = 7 x 0.02s =0.14s
Average Velocity, v = 12 cm / 0.14s
= 85.71 cms-1
-
8/6/2019 Analysing of Motion1
15/57
Examples : Determine the acceleration
8
7
6
5
4
3
2
1
0
u
vLength / cm
Ticks
L1
L2
-
8/6/2019 Analysing of Motion1
16/57
Figure above shows a chart formed from strips
of ticker tape with ten ticks.
a) The time foreach 10-tick strip
= 10 x 0.02 = 0.2s
b) The average initial velocity, u
u = L1/ 0.2
c) The average final velocity, v
v = L2 / 0.2
-
8/6/2019 Analysing of Motion1
17/57
d) The time, t taken between u and v can be
determined by the following:
t = (total number of strips x (Time for
of tape between u &v) each strip)
= 5 x 0.2
=1.0s
-
8/6/2019 Analysing of Motion1
18/57
-
8/6/2019 Analysing of Motion1
19/57
-
8/6/2019 Analysing of Motion1
20/57
-
8/6/2019 Analysing of Motion1
21/57
-
8/6/2019 Analysing of Motion1
22/57
Equation ofLinear Motion
-
8/6/2019 Analysing of Motion1
23/57
Learning Outcomes
By theend of the lesson, you should be
able to:
Solve problems on linear motion withuniform acceleration using
i. V = u + at
ii. S= ut + at2
iii. V2= u2 + 2as
-
8/6/2019 Analysing of Motion1
24/57
Equation of Motion
v2 = u2 + 2as
v = u + at
s = ut + at2
a = v u
t
s = v t
tvus v
!
2
Examples
-
8/6/2019 Analysing of Motion1
25/57
-
8/6/2019 Analysing of Motion1
26/57
-
8/6/2019 Analysing of Motion1
27/57
-
8/6/2019 Analysing of Motion1
28/57
Average velocity, v = u + v
2
s = u + v
t 2
tvu
s v
!
2
-
8/6/2019 Analysing of Motion1
29/57
Examples
A car is travelling with a uniform velocity of
80 km h-1 northward from Johor Bahru. What
is its displacement after15 minutes?
-
8/6/2019 Analysing of Motion1
30/57
Time = 15 minutes = 0.25 hour
Using the formula, s = v x t,
S = 80 km h-1
x 0.25 h = 20 kmTherefore, its displacement is 20km north
of Johor Bahru.
-
8/6/2019 Analysing of Motion1
31/57
Examples
A school bus accelerates with an
acceleration of4.0 ms-2 after picking up
some students at a bus stop.
Calculate the
a) velocity
b) distance
travelled by the bus after 5 s.
-
8/6/2019 Analysing of Motion1
32/57
-
8/6/2019 Analysing of Motion1
33/57
b) S = ut + at2
= (0 x 5) + ( x 4.0 x 52)
= 50 m
-
8/6/2019 Analysing of Motion1
34/57
Examples
Jerak was running at 2.5 ms-1
towards thelift. When he was 15 m away from the lift,
the door of the lift was due to close
completely in 5 s.
a) Explain why Jerak did not manage to enter
the lift.
b) In order to be able to enter the lift before
its door closed completely, Jerak needed
to accelerate. Calculate the minimum
acceleration needed by Jerak to do so.
-
8/6/2019 Analysing of Motion1
35/57
a) When Jerak ran at a constant velocity of
2.5 ms-1, the distance, s travelled by him
in 5s is:
s = v x t = 2.5 x 5 = 12.5 m
Since he was 15m away from the lift, he
did not manage to reach the lift before its
door closed.
-
8/6/2019 Analysing of Motion1
36/57
b) Let the minimum acceleration needed byJerak to reach the lift bea.
Then heentered the lift after 5 s, the time
interval before the door of the lift closedcompletely.
S = ut + at2
s = 15m; t = 5s; u = 2.5 ms-1
15 = (2.5x5) + ( a x 52)
Therefore, a = 0.2 ms-2
-
8/6/2019 Analysing of Motion1
37/57
-
8/6/2019 Analysing of Motion1
38/57
a) v2= u2+ 2as
a = 8.3 ms-1
b)
t = 0.6 s
tvu
s v
!
2
-
8/6/2019 Analysing of Motion1
39/57
Exercises
-
8/6/2019 Analysing of Motion1
40/57
1. A cyclist accelerates uniformly at 1.2 ms-
1 in 10s from rest. What is his
displacement at this time?
-
8/6/2019 Analysing of Motion1
41/57
-
8/6/2019 Analysing of Motion1
42/57
2. A lorry approaches a traffic light at 15 ms-1.
The light turns red and the driver applies
the brakes at a distance of25m from the
junction and accelerates a -5 ms-2. Can thelorry come to a complete stop before
reaching the stop line at the traffic light?
-
8/6/2019 Analysing of Motion1
43/57
-
8/6/2019 Analysing of Motion1
44/57
3. A cargo ship sails across the Straits of
Malacca towards Klang North Port.
Within a given time, it can accelerate
0.25 ms-2in 52 s from its initial velocity of
2 ms-1. Determine its final velocity at the
end of the accelerating period.
-
8/6/2019 Analysing of Motion1
45/57
-
8/6/2019 Analysing of Motion1
46/57
4. A car accelerates from 10 ms-1 to 50 ms-1
in 20 seconds. What is its acceleration?
-
8/6/2019 Analysing of Motion1
47/57
Given the information,Initial velocity, u = 10 ms-1
Final velocity, v = 50 ms-1
Time, t = 20 s
a = (v-u)/t
= 50ms-1 10 ms-1
20s
= 2 ms-2
-
8/6/2019 Analysing of Motion1
48/57
5. A car travels from Penang to Ipoh at an
average speed of100 km h-1. If the
distance between Penang and Ipoh is
200km, how long does it take to travel to
Ipoh?
-
8/6/2019 Analysing of Motion1
49/57
Given the information,
Average speed, v = 100 km h-1
Distance, s = 200 km
t = ?
s = v x t
t = s/v
= 200 km /100 km h-1
= 2 h
-
8/6/2019 Analysing of Motion1
50/57
Activity
-
8/6/2019 Analysing of Motion1
51/57
Answer a)
-
8/6/2019 Analysing of Motion1
52/57
-
8/6/2019 Analysing of Motion1
53/57
Answer b)
-
8/6/2019 Analysing of Motion1
54/57
-
8/6/2019 Analysing of Motion1
55/57
Answer c)
-
8/6/2019 Analysing of Motion1
56/57
-
8/6/2019 Analysing of Motion1
57/57