PHY130 – FUNDAMENTAL PHYSICS I CHAPTER 2 KINEMATICS IN ONE DIRECTION
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Lesson Contents
2.1 Scalars and vectors 2.2 Linear motion parameters 2.2.1 Definition of linear motion parameters 2.2.2 Average and instantaneous velocity 2.2.3 Average and instantaneous acceleration
2.3 Graph of linear motion 2.3.1 Displacement – time graph
2.3.2 Velocity – time graph 2.4 Linear motion with constant acceleration
2.5 Free fall motion
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Lesson Outcomes
After studying this chapter, you should : • Able to define & write equation for the motion parameter: displacement, velocity and acceleration
• Able to explain & differentiate between the average & instantaneous values of the velocity and acceleration • Able to determine the motion parameters from the displacement-time graph and the velocity-time graph • Able to explain and analyze motion with constant acceleration • Able to explain, analyze and solve falling object problems 3
2.1 SCALARS AND VECTORS
Scalar & Vectors
Each physical quantity can be categorized as either a scalar quantity or a vector quantity.
QUANTITIES
Scalar
Vector
DEFINITION
physical quantities that have magnitude only
physical quantities that possess both magnitude and direction.
EXAMPLES
• Distance • Work • Charge • Speed • Power
• Displaceme • Momentum nt • Velocity • Electric field • Force
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2.2 LINEAR MOTION PARAMETERS
Distance & Displacement
DISTANCE
DISPLACEMENT
DEFINITION
total path length the shortest distance from the travelled from one initial position to the final position location to another
QUANTITY
scalar
vector
SI UNIT
metre (m)
metre (m)
VALUE
Always positive
1. Either Positive, negative or zero • Positive : right ( ) • Negative : left / opposite ( ) 2. In calculating displacement = Final position – Initial position
x = xf - xi 7
Speed & Velocity
SPEED
VELOCITY
DEFINITION
How fast an object travels in a given time interval
rate of change of displacement
QUANTITY
Scalar
Vector
SI UNIT
m/s @ ms-1
m/s @ ms-1
FORMULA
Distance / Time
Displacement / Time interval **direction of velocity = direction of change in displacement
TERMS
Instantaneous Speed : Speed at any instant
Instantaneous velocity : Velocity at any instant
Constant / uniform speed : Equal distance moved in equal time intervals
Constant / uniform velocity : Velocity does not increase or slow down
Average speed 
total distance moved total time taken
Average velocity 
total displacement total time taken 8
Acceleration
ACCELERATION DEFINITION
A change of velocity over time interval
QUANTITY
Vector
UNIT
m/s2 @ ms-2
TERMS
Deceleration = Negative acceleration
Instantaneous acceleration : Acceleration at any instant. constant acceleration : velocity increases @ decreases by the same amount for every second of its motion.
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s
Relationship of velocity and displacement
Gradient = constant speed = gradient
0
(a) Uniform velocity
t
s
s Q
(c)
P
R
Gradient at point R is negative.
Gradient at point Q is zero.
0
t
Gradient increases with time
The direction of velocity is changing.
The velocity is zero.
0 (b) The velocity increasest with time
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Relationship of velocity and acceleration
Area under v-t graph =
displacement/distance Gradient = acceleration
v
v
v
B
Uniform acceleration
Uniform velocity
C
A 0
t1 (a) t2
t
0
t1
(b) t2
t
0
t1
t2 (c)
t
2.3 GRAPHS OF LINEAR MOTION
EXAMPLE A toy train moves slowly along a straight track according to the displacement, s against time, t graph in figure below.
s (cm)
(i) Displacement - time graph
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8 6 4 2 0
2
4
6
8
10 12 14
t (s)
TIPS!!You must know to: 1. Sketch a graph from the given information or vice versa 2. Calculate the gradient (velocity/speed), average velocity and instantaneous velocity
a. Sketch a velocity (cm s-1) against time (s) graph. b. Determine the average velocity for the whole journey.
c. Calculate the instantaneous velocity at t = 12 s.
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SOLUTION a)
v (cm s1)
s2 s1 b) vav t 2 t1
1.50
0.68 0
2
4
6
8
10 12 14
t (s)
10 0 vav 14 0 v a v 0 .7 1 4 c m s 1
c)v average velocity from 10 s to 14 s s2 s1 10 4 v t2 t1 14 10
1 .5 0 c m s
1
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EXAMPLE A velocity-time (v-t) graph in figure below shows the motion of a lift.
v (m s ď€1) (ii) Velocity - time graph
4 2 0 -2 -4
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10 15
20 25 30 35 40 45
50
t (s)
TIPS!!You must know to: 1. Sketch a graph from the given information or vice versa 2. Calculate the gradient (acceleration) or area of graph (displacement/distance)
a. Sketch a graph of acceleration (m s-2) against time (s). b. Determine the total distance traveled by the lift and its displacement. c. Calculate the average acceleration between 20 s to 40 s.
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SOLUTION a (m sï€2)
a)
a.
0.8 0.6 0.4 0.2 0 -0.2
5
10 15
20 25 30 35 40 45
50
t (s)
-0.4
-0.6 -0.8 16
SOLUTION b)i
TIPS!! 1.Total distance = total area of the graph (both + Y axis and –Y axis)
v (m s 1) 4 2 0 -2
A1 5
A2 10 15
A3
2. Total displacement • area at + Y axis = +ve area • Area at –Y axis = -ve area
20 25 30 A 35 40 45 4 A5
50
t (s)
-4
Total distance area under the graph of v-t A1 A 2 A3 A 4 A5 1 1 1 1 1 Total distance 25 2 410 5 104 54 15 54 2 2 2 2 2
T o ta l d is ta n c e 1 1 5 m
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SOLUTION Displacement area under the graph of v-t
b) ii
A1 A 2 A3 A 4 A5
c)
1 25 1 2 410 1 5 104 1 5 4 1 15 5 4 2 2 2 2 2
D is p la c e m e n t 1 5 m v2 v1 aav t 2 t1 44 aav 40 20 a a v 0 .4 m s 2
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2.4 LINEAR MOTION WITH CONSTANT ACCELERATION
REMEMBER!!
The 4 equations
TIPS!! Where
v = u + at s = ut + 2
1 2
at
2
2
v = u + 2as s=
1 2
(u + v)t
v = final velocity u = initial velocity a = acceleration t = time s = displacement
JUES
Word Symbol numbers 1. Read the question carefully. 2. JOT DOWN the given information using symbols and value. Eg : acc, a = 10 ms-2 3. What is the unknown value? HINT WORD : calculate, determine. 4. Write in symbols. Eg : final velocity, v= ?? 5. Select the most suitable equation 6. Solve it. GOOD LUCK!!
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EXAMPLE 1. A box starts from rest and moves with a speed of 2.2 ms-1 in 3 seconds. Calculate the : [MARCH 2014] (i)
acceleration of the box
(ii) distance it moved in the first 6 seconds.
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SOLUTION i)
Acceleration of the box v = u + at
s = ut +
2.2 = 0 + a(3) a = 0.733 ms
ii) Distance it
s =0+ -2
1 2 1
2
moved in the first 6 seconds.
at
2
(0.733)(6)
s = 13.194 m
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2.5 FREE FALL MOTION
1) 2)
The direction of motion are now along the vertical y-axis. Vector quantities (displacement, initial velocity & final velocity) are positive in an upward motion & negative in a downward motion. + s y, + u y, + v y reference level
- s y, - u y, - v y
3)
The free-fall acceleration is always negative in an upward & a downward motion. - g = - 9.81 m s
-2
- g = - 9.81 m s
-2
reference level
4)
t is always positive.
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TRANSITION
PREVIOUS
The 4 equations
Where
v = final velocity u = initial velocity a = acceleration g = gravitational acceleration t = time s = displacement
v = u + at s = ut + 2
1 2
at
2
2
v = u + 2as s=
1 2
NOW
TIPS!! 1) a ďƒ -g = - 9.81 ms-2
v = u - gt s = ut -
2) Max height : v = 0 ms-1 3) Reference point : s = 0 m
2
+ s y, + u y, + v y, + t reference level
2
gt
2
2
v = u - 2gs
4)
(u + v)t
1
s=
1 2
(u + v)t
- s y, - u y, - v y, + t 25
EXAMPLE A book is dropped 150 m from the ground. Determine a. the time taken for the book reaches the ground. b. the velocity of the book when it reaches the ground. (given g = 9.81 m s-2)
uy = 0 m s1 Hence 150 m
1 2 s y u y t gt 2 1 150 0 9.81t 2 2
t 5.53 s
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SOLUTION a. The vertical displacement is sy = 150 m
1 2 s y u y t gt 2 1 150 0 9.81t 2 2 t 5.53 s
SOLUTION b. The book’s velocity is given by
uy 0
vy u y gt
v y 0 9.81 5.53 1 v y 54.2 m s OR
s y 150 m
v y u y 2 gs y 2
2
v y 0 2 9.81 150 2
v y ? v 54.2 m s 1 ( downward ) y Therefore the book’s velocity is
v y 54.2 m s
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