Motion Graph & equations
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A student is able to: - Plot and interpret displacement-time and velocity-time graphs - Determine an object's motion from the shape of the graphs, including whether it is at rest, moving with uniform or non-uniform velocity/acceleration - Calculate distance, displacement, velocity, and acceleration from displacement-time and velocity-time graphs - Solve problems involving linear motion using the equations of motion
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Derived the equations
Acceleration = Gradient of
v-t graph
Rearrange [1]
Final velocity v = u+at …….[1a]
Displacement, s = area under v-t
graph
s = 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](https://image.slidesharecdn.com/2-170404135342/85/Motion-Graph-equations-28-320.jpg)
Motion Graph & equations
- 1. 1
- 2. 2 •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 . A student is able to: Learning OutcomesLearning Outcomes
- 3. 3 MOTION GRAPHS DISPLACEMENT – TIME GRAPH VELOCITY – TIME GRAPH GRADIENT : VELOCITY (v) 1: GRADIENT : ACCELERATION (a) 2: AREA UNDER THE GRAPH : DISPLACEMENT (s)
- 4. 4 DISPLACEMENT – TIME GRAPHDISPLACEMENT – TIME GRAPH s / m t /s Gradient = s t s t = velocity = m s = ms-1
- 5. 55 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 Example 1Example 1
- 6. 6
- 7. 77 s : constant v : zero s t GRAPH 1GRAPH 1
- 8. 88 GRAPH 2GRAPH 2 s t s increases uniformly v uniform
- 9. 99 GRAPH 3GRAPH 3 Larger increment in s v increases s t
- 10. 1010 s t GRAPH 4GRAPH 4 Smaller increment in s v decreases
- 11. 11 GRAPH 5GRAPH 5 s t -v uniform •Object moves at opposite direction s decreases uniformly
- 12. 12 GRAPH 6GRAPH 6 s t Larger decrement in s -v ?
- 13. 1313 GRAPH 7GRAPH 7 s t -v ? Smaller decrement in s
- 14. 14 GRAPH VELOCITY - TIMEGRAPH VELOCITY - TIME Gradient = v - u t v / ms-1 t /s v - u t = ms-1 s = ms-2 = acceleration
- 15. 1515 Example 2Example 2 v / 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
- 16. 16 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
- 17. 17 v / ms-1 t /st v Area under the graph = rectangle area = ( v )(t) GRAPH VELOCITY - TIMEGRAPH VELOCITY - TIME
- 18. 18 ExampleExample 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 + area =½ (8-2)m/s(4 s) + (2 m/s)(4s) = 20 m
- 19. 19
- 20. 2020 v : constant a : zero v t GRAPH 1GRAPH 1
- 21. 2121 GRAPH 2GRAPH 2 v t v increases uniformly a uniform
- 22. 2222 GRAPH 3GRAPH 3 Larger increment in v a increases v t
- 23. 2323 v t GRAPH 4GRAPH 4 Smaller increment in v a decreases
- 24. 24 GRAPH 5GRAPH 5 v t -a uniform v decreases uniformly
- 25. 25 GRAPH 6GRAPH 6 v t Larger decrement in v -a ?
- 26. 2626 GRAPH 7GRAPH 7 v t -a ? Smaller decrement in v
- 27. 27 Equations of Linear MotionEquations of Linear Motion Derived the equations Problem solving
- 28. 28 Derived the equations Acceleration = Gradient of v-t graph Rearrange [1] Final velocity v = u+at …….[1a] Displacement, s = area under v-t graph s = 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
- 29. 29 ConclusionConclusion EquationEquation vv ss tt uu vv aa 11 22 33 44 55 66 a = (v – u) t v = s t s = ut +½ at2 v2 = u2 + 2as v = (v + u) 2 s = ½ ( u+v) t √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √
- 30. 30 Where s = displacement u = Initial velocity v = Final velocity a = Constant acceleration t = time interval The Equations of Motion atuv +=1. u+v 2 S = t4. asuv 222 +=3. 2 2 1 atuts +=2.
- 31. 31 Example 1Example 1 A train is moving at a constant velocity ofA train is moving at a constant velocity of 10 m/s,. It is then uniformly accelerated in10 m/s,. It is then uniformly accelerated in 10 s to a velocity of 20 m/s. what is the10 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
- 32. 32 Example 2Example 2 A cheetah accelerates from rest at 4 m/sA cheetah accelerates from rest at 4 m/s22 over a distance of 60m to catch a rabbit.over a distance of 60m to catch a rabbit. What is the final speed?What is the final speed? u = 0 m/s a = 4 m/s2 s = 60 m v2 = u2 + 2as
- 33. 33 Example 3Example 3 A new car start from rest and travel 400m in 16 s.A new car start from rest and travel 400m in 16 s. a) What is its average acceleration duringa) 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
- 34. 34