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•plot and interpret displacement- time and velocity-time graphs•deduce from the shape of a displacement-time graph when a body is: i. at rest ii. moving with uniform velocity iii. moving with non-uniform velocity •determine distance, displacement and velocity from a displacement –time graph•deduce from the shape of velocity- time graph when a body is: i. at rest ii. moving with uniform velocity iii. moving with uniform acceleration •determine distance, displacement velocity and acceleration from a velocity–time graph•solve problems on linear motion with uniform

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A student is able to:Learning OutcomesLearning Outcomes

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MOTION GRAPHS

DISPLACEMENT – TIME GRAPH

VELOCITY – TIME GRAPH

GRADIENT : VELOCITY (v)

1: GRADIENT : ACCELERATION (a)

2: AREA UNDER THE GRAPH : DISPLACEMENT (s)

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DISPLACEMENT – TIME GRAPHDISPLACEMENT – TIME GRAPHs / m

t /s

Gradient = s t

s

t= velocity

= m s= ms-1

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s / cm

t / s

10

2

v = gradient of s-t graph = (10 – 0 ) cm ( 2 – 0 ) s = 5 cm/s

Base on the graph, find the velocity of the object ‘s movement.

Example 1 Example 1

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77

s : constant

v : zero

s

t

GRAPH 1GRAPH 1

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GRAPH 2GRAPH 2

s

t

s increases uniformly

v uniform

99

GRAPH 3GRAPH 3

Larger increment in s

v increases

s

t

1010

s

t

GRAPH 4GRAPH 4

Smaller increment in s

v decreases

1111

GRAPH 5GRAPH 5

s

t

-v uniform •Object moves at opposite direction

s decreases uniformly

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GRAPH 6GRAPH 6

s

t

Larger decrement in s

-v ?

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GRAPH 7GRAPH 7

s

t

-v ?

Smaller decrement in s

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GRAPH VELOCITY - TIMEGRAPH VELOCITY - TIME

Gradient = v - u t

v / ms-1

t /s

v - u

t

= ms-1

s= ms-2

= acceleration

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Example 2Example 2v / cms-1

t / s

10

2

a = gradient of v-t graph = (10 – 0 ) cms-1 ( 2 – 0 ) s = 5 cm/s-2

Base on the graph, find the acceleration of the object ‘s movement.

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Area under the graph = triangle area = 1 ( v )(t)

2

= (ms-1)(s)

v / ms-1

t /s

v

t

= displacement

= m

GRAPH VELOCITY - TIMEGRAPH VELOCITY - TIME

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v / ms-1

t /st

v

Area under the graph = rectangle area = ( v )(t)

GRAPH VELOCITY - TIMEGRAPH VELOCITY - TIME

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Example Example v (m/s)

t(s)

2

4

8

Base on the graph, calculate the displacement of a moving object.

Answer

Displacement = Area under v-t graph

= ∆ area + ٱarea

=½ (8-2)m/s(4 s) + (2 m/s)(4s)

= 20 m

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2020

v : constant

a : zero

v

t

GRAPH 1GRAPH 1

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GRAPH 2GRAPH 2

v

t

v increases uniformly

a uniform

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GRAPH 3GRAPH 3

Larger increment in v

a increases

v

t

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v

t

GRAPH 4GRAPH 4

Smaller increment in v

a decreases

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GRAPH 5GRAPH 5

v

t

-a uniform

v decreases uniformly

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GRAPH 6GRAPH 6

v

t

Larger decrement in v

-a ?

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GRAPH 7GRAPH 7

v

t

-a ?

Smaller decrement in v

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Equations of Linear MotionEquations of Linear Motion

Derived the equations

Problem solving

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Derived the equationsAcceleration = Gradient of v-t graph

Rearrange [1]

Final velocity v = u+at …….[1a]

Displacement, s = area under v-t graphs = ½ ( u+v) t ……..[2]

Substitute [1a] into [2]

velocity

time

u

t

v

s = ½ ( u+ u+at) t

s = ut +½ at2……...[3]

Substitute [1b] into [2]

Rearrange [1] again

s = ½ ( u+v) (v – u) a

v2 = u2 + 2as …….[4]

t = (v – u) ……..[1b] a

a = (v – u) ……..[1] t

v = (v + u) ……..[5] 2

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Conclusion Conclusion Equation Equation v v ss tt uu vv aa

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22

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445566

a = (v – u) t

v = s t

s = ut +½ at2

v2 = u2 + 2as

v = (v + u) 2

s = ½ ( u+v) t

√ √ √

√ √ √

√ √ √ √

√ √ √ √

√ √ √ √

√ √ √ √

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Wheres = displacementu = Initial velocityv = Final velocitya = Constant accelerationt = time interval

The Equations of Motion

atuv 1.

u v2

S = t4.

asuv 222 3.

2

21 atuts 2.

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Example 1Example 1 A train is moving at a constant velocity of A train is moving at a constant velocity of

10 m/s,. It is then uniformly accelerated in 10 m/s,. It is then uniformly accelerated in 10 s to a velocity of 20 m/s. what is the 10 s to a velocity of 20 m/s. what is the distance traveled by the train?distance traveled by the train?

u = 10 m/s

v = 20 m/s

t = 10 s

s = (v + u) t 2 = (20 – 10)m/s(10s)

2

s = 50 m

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Example 2Example 2 A cheetah accelerates from rest at 4 A cheetah accelerates from rest at 4

m/sm/s22 over a distance of 60m to catch over a distance of 60m to catch a rabbit. What is the final speed?a rabbit. What is the final speed?

u = 0 m/s

a = 4 m/s2

s = 60 m

v2 = u2 + 2as

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Example 3Example 3 A new car start from rest and travel 400m in 16 A new car start from rest and travel 400m in 16

s.s. a) What is its average acceleration during a) What is its average acceleration during this time?this time? b) Calculate the final speed of the car.b) Calculate the final speed of the car. c) How fast is this final speed in kmhc) How fast is this final speed in kmh-1-1

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