physics Linear Motion

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CHAPTER 2

FORCES AND MOTION

2.1 Analysing Linear Motion

Expected learning outcomes

Student should be able to :

1. define distance and displacement

2. define speed and velocity

3. define acceleration and deceleration

4. calculate speed and velocity

5. calculate acceleration/deceleration

6. solve problems on linear motion with uniform acceleration using :

(i) v = u + at

(ii) s = ut + at2

(iii) v2= 2as


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Distance and Displacement

1. Distance is the total path length travelled from one location toanother. It is a scalarquantity.The SI Unit is meter (m)

2.Displacement is the distance between two locations measured

along the shortest path connecting them, in a specified direction.

It is a vector quantity. The SI Unit is meter(m)


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2m 2 m Distance = 4 m

Displacement = 4 m

3m1 m

Distance = 3 + 1 +1 = 5m

Displacement = 3m + 1+ ( -1) =3 m toright

A B

AC

B

A B C


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Exercise 1:

1. What is the distance he had to drive to reach his school?

2. What is the displacement of the school from the house?

3. What is the total distance he had to drive to reach back to

his house?

4. What is the total displacement he drove back to his

house?


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Speed and Velocity

Speed is the distance travelled per unit time or the rate of change ofdistance.

Average speed, v= Total distance travelled, s (m) SI unit = m/s

Time taken, t(s)

Velocity is the speed in a given direction or the rate of change ofdisplacement.

Average velocity, v=Displacement, s (m)

Time taken, t(s)


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Q1 : If the time taken by Ahmad to travel from his house to school

is 15 minutes, what is his speed and velocity?


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Positive velocity

Displacement : Right directionDisplacement left direction

Negative velocity


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Acceleration and Deceleration1. Acceleration is defined as the rate of change of velocity.

Acceleration = Change of velocity

Time taken

= Final velocity Initial velocity

Time taken

a = vu

t

The acceleration of an object is a vector quantity. The SI units for speed

is ms-2

Negative acceleration is known as deceleration. If the velocity of an

object is remain constant, the object is said to move with zero

acceleration.

If the rate of change of its velocity is constant, the object move uniformacceleration or zero acceleration.


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Acceleration, a = vu

t

A runner accelerates at a constant rate from rest and reaches a

velocity of 10 ms-1 after 5.0s. What is his acceleration?

Initial velocity, u = 0

Final velocity, v = 10ms-1

Time interval, t = 5.0s= 10 - 0

5= 2 ms-2

1. Calculate its average acceleration.

2. A motorcycle, travelling at 40.0 ms-1, takes

10s to stop. What is its average deceleration?


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A car decrease its velocity from 108 kmh-1 to 72 kmh-1 in 5 s. What is the

deceleration of the car ?

Solution :


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Next Click

>play


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The time taken to make 50 dots on the ticker tape is 1s. Hence , the

time interval between 2 consecutive dots (tick) is 1/50 = 0.02s.


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tick

Strip

5 tick strip

tick tick tick tick

1 2 3 4 5

A tape chart


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Direction of the movement of ticker tape

Direction of the movement of ticker tape

v

t


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Direction of the movement of the ticker tape

v

t


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v

t

Direction of the movement of the ticker tape


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To determine the velocity of a ticker tape

30.0 cm

Velocity = Displacement

Time taken

v = s

t

s = 30

7x

0.02

V =

214.28m/s


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To determine the velocity of strips of ticker tape

velocity

v = s

t

v = 8+6+9+8+10

1.0

v = 41.0 cms-1


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To determine acceleration from the ticker tape

=833.33cm/s


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Ticker tape chart

Length/cm

time

Physics is fun

We can determine

velocity,

acceleration,time

taken and

acceleration from

ticker tape chart!!!


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The Equation of Motion

s = Displacementu= Initial velocity

v= Final velocity

a = Constant acceleration

t= Time interval

V = u+ ats= ut+ at2

v2 = u2 + 2as

V = s/t

s = u+ v t

2


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Hazuri is driving a car at a velocity of 30 ms-1. On seeing a student

crossing the road, Hazuri steps on his brakes to stop the car. The

speed of the car decreases uniformly and comes to a halt after

travelling 150m.a) What is the deceleration of the car when the brakes are applied.

b) What is the time interval before the car stops.


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Try your understanding!!

1. Jerry rides his bicycle starting from restand accelerates at a uniform rate of0.5ms-2. What is the time taken to travel a

distance of 100m? 2. The velocity of a car, decelerating

uniformly, decreases from 10ms to 6 ms.The distance travelled in the period ofdeceleration is 16m. What is the additionaldistance travelled before the car stops?

-1 -1


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Check your answer

1. s = ut + a t

100 = 0 + (0.5) t2

t = 400 s

t = 20 s

2. v =u + 2 as

62=102 + 2 a (16)

a =- 2 ms

Whenthe bicyclestops,v= 0

U = 6 ms v = 0 ms-2

v2=u2 + 2 as 02=62 + 2 (-2) s

4s=36

.s =9 m

Good try!!

2

2

2 2

-2


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Right hand>.play


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Right hand>.play


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Right hand>.play

physics Linear Motion

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