Analysing of Motion1

8/6/2019 Analysing of Motion1

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Analysing of Motion


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Learning Outcomes

Determine displacement, average velocity

and acceleration.


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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.


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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.


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1.

The distance between two neighbouring

dots are the same throughtout the tape.

Therefore the trolley moved with uniform

velocity.


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2.

The distances between two neighbouring dotsare increasing.

Therefore the trolley moved with increasing

velocity.

The trolley was accelerating.


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3.

The distances between two neighbouring

dots are decreasing.

Therefore, the trolley moved with decreasing

velocity.

The trolley was decelerating.


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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

. . . . . . . .


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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


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Examples : Determine the acceleration

8

7

6

5

4

3

2

1

0

u

vLength / cm

Ticks

L1

L2


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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


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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


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Equation ofLinear Motion


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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


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Equation of Motion

v2 = u2 + 2as

v = u + at

s = ut + at2

a = v u

t

s = v t

tvus v

!

2

Examples


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Average velocity, v = u + v

2

s = u + v

t 2

tvu

s v

!

2


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Examples

A car is travelling with a uniform velocity of

80 km h-1 northward from Johor Bahru. What

is its displacement after15 minutes?


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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.


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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.


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b) S = ut + at2

= (0 x 5) + ( x 4.0 x 52)

= 50 m


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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.


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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.


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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


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a) v2= u2+ 2as

a = 8.3 ms-1

b)

t = 0.6 s

tvu

s v

!

2


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Exercises


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1. A cyclist accelerates uniformly at 1.2 ms-

1 in 10s from rest. What is his

displacement at this time?


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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?


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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.


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4. A car accelerates from 10 ms-1 to 50 ms-1

in 20 seconds. What is its acceleration?


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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


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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?


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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


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Activity


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Answer a)


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Answer b)


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Answer c)


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