Post on 06-Dec-2015
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Kinematics definitions
• Kinematics – branch of physics; study of motion
• Position (x) – where you are located• Distance (d ) – how far you have traveled,
regardless of direction • Displacement (x) – where you are in
relation to where you started
Kinematics of Linear motion
• Defined as the studies of motion of an objects without considering the effects that produce the motion.
• There are two types of motion:– Linear or straight line motion (1-D)
• with constant (uniform) velocity• with constant (uniform) acceleration, e.g. free fall motion
– Projectile motion (2-D)• x-component (horizontal)• y-component (vertical)
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Linear motion (1-D)
Distance, d
• scalar quantity.• is defined as the length of actual path between
two points.• For example :
– The length of the path from P to Q is 25 cm. 4
P
Q
s
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vector quantity. is defined as the distance between initial point and final point in
a straight line. The S.I. unit of displacement is metre (m).
Example 2.1 :An object P moves 30 m to the east after that 15 m to the south
and finally moves 40 m to west. Determine the displacement of P
relative to the original position.
Solution :
Displacement,
N
EW
S
O
P
30 m
15 m
10 m 30 m
Distance vs. Displacement• You drive the path, and your odometer goes
up by 8 miles (your distance).• Your displacement is the shorter directed
distance from start to stop (green arrow).• What if you drove in a circle?
start
stop
The magnitude of the displacement is given by
and its direction is
2.1.3 Speed, v
• is defined as the rate of change of distance.• scalar quantity.• Equation: interval time
distance of changespeed
7
Δt
Δdv
m 181015 22 OP
south west tofrom 5610
15tan 1
θ
Speed, Velocity, & Acceleration
• Speed (v) – how fast you go
• Velocity (v) – how fast and which way; the rate at which position changes
• Average speed ( v ) – distance / time
• Acceleration (a) – how fast you speed up, slow down, or change direction; the rate at which velocity changes
v
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interval time
ntdisplaceme of changeavv
12
12av tt
ssv
is a vector quantity. The S.I. unit for velocity is m s-1.
Average velocity, vav
is defined as the rate of change of displacement. Equation:
Its direction is in the same direction of the change in displacement.
Velocity,
Δt
Δsvav
constantdt
ds
10
t
s
0tv
limit
Instantaneous velocity, v is defined as the instantaneous rate of change of displacement. Equation:
An object moves in a uniform velocity when
and the instantaneous velocity equals to the average velocity at any time.
dt
dsv
Speed vs. Velocity
• Speed is a scalar (how fast something is moving regardless of its direction). Ex: v = 20 mph
• Speed is the magnitude of velocity.• Velocity is a combination of speed and direction. Ex:
v = 20 mph at 15 south of west
• The symbol for speed is v.• The symbol for velocity is type written in bold: v or
hand written with an arrow: v
Speed vs. Velocity
• During your 8 mi. trip, which took 15 min., your speedometer displays your instantaneous speed, which varies throughout the trip.
• Your average speed is 32 mi/hr.• Your average velocity is 32 mi/hr in a SE direction.• At any point in time, your velocity vector points
tangent to your path. • The faster you go, the longer your velocity vector.
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Therefore
Q
s
t0
s1
t1
The gradient of the tangent to the curve at point Q
= the instantaneous velocity at time, t = t1
Gradient of s-t graph = velocity
a
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interval time
velocityof changeava
vector quantity. The S.I. unit for acceleration is m s-2.
Average acceleration, aav
is defined as the rate of change of velocity. Equation:
Its direction is in the same direction of motion. The acceleration of an object is uniform when the magnitude of
velocity changes at a constant rate and along fixed direction.
Acceleration,
12
12av tt
vva
Δt
Δvaav
AccelerationAcceleration – how fast you speed up, slow
down, or change direction; it’s the rate at which velocity changes. Two examples:
t (s) v (mph)
0 55
1 57
2 59
3 61
t (s) v (m/s)
0 34
1 31
2 28
3 25
a = +2 mph / s a = -3 m / ss = -3 m / s
2
constantdt
dv
16
t
v
0ta
limit
Instantaneous acceleration, a is defined as the instantaneous rate of change of velocity. Equation:
An object moves in a uniform acceleration when
and the instantaneous acceleration equals to the average acceleration at any time.
2
2
dt
sd
dt
dva
Velocity & Acceleration Sign Chart
V E L O C I T Y ACCELERATION
+ -
+ Moving forward;
Speeding up
Moving backward;
Slowing down
- Moving forward;
Slowing down
Moving backward;
Speeding up
Acceleration due to Gravity
9.81 m/s2
Near the surface of the Earth, all objects accelerate at the same rate (ignoring air resistance).
a = -g = -9.81 m/s2
Interpretation: Velocity decreases by 9.81 m/s each second, meaning velocity is becoming less positive or more negative. Less positive means slowing down while going up. More negative means speeding up while going down.
This acceleration vector is the same on the way up, at the top, and on the way down!
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Deceleration, a is a negative acceleration. The object is slowing down meaning the speed of the object
decreases with time.
Therefore
v
t
Q
0
v1
t1
The gradient of the tangent to the curve at point Q
= the instantaneous acceleration at time, t = t1
Gradient of v-t graph = acceleration
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Displacement against time graph (s-t)Graphical methods
s
t0
s
t0(a) Uniform velocity (b) The velocity increases with time
Gradient = constant
Gradient increases with time
(c)
s
t0
Q
RP
The direction of velocity is changing.
Gradient at point R is negative.
Gradient at point Q is zero.
The velocity is zero.
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Velocity versus time graph (v-t)
The gradient at point A is positive – a > 0(speeding up) The gradient at point B is zero – a= 0 The gradient at point C is negative – a < 0(slowing down)
t1 t2
v
t0 (a) t2t1
v
t0(b) t1 t2
v
t0(c)
Uniform velocity
Uniform acceleration
Area under the v-t graph = displacement
BC
A
dt
dsv
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From the equation of instantaneous velocity,
Therefore
vdtds
2
1
t
tvdts
graph under the area dedsha tvs
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A toy train moves slowly along a straight track according to the
displacement, s against time, t graph in figure below.
a. Explain qualitatively the motion of the toy train.
b. Sketch a velocity (cm s-1) against time (s) graph.
c. Determine the average velocity for the whole journey.
d. Calculate the instantaneous velocity at t = 12 s.
e. Determine the distance travelled by the toy train.
Example :
0 2 4 6 8 10 12 14 t (s)
2
4
6
8
10
s (cm)
25
Solution :
a. 0 to 6 s : The train moves at a constant velocity of
6 to 10 s : The train stops.
10 to 14 s : The train moves in the same direction at a constant velocity of
b.
0 2 4 6 8 10 12 14 t (s)
0.68
1.50
v (cm s1)
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Solution :
c.
d.
e. The distance travelled by the toy train is 10 cm.
12
12
tt
ssvav
s 14 tos 10 from velocity averagev
12
12
tt
ssv
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A velocity-time (v-t) graph in figure below shows the motion of a lift.
a. Describe qualitatively the motion of the lift.
b. Sketch a graph of acceleration (m s2) against time (s).
c. Determine the total distance travelled by the lift and its
displacement.
d. Calculate the average acceleration between 20 s to 40 s.
Example :
05 10 15 20 25 30 35 t (s)
-4
-2
2
4
v (m s1)
40 45 50
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Solution :
a. 0 to 5 s : Lift moves upward from rest with a constant acceleration of
5 to 15 s : The velocity of the lift increases from 2 m s1 to 4 m s1 but the acceleration decreasing to
15 to 20 s : Lift moving with constant velocity of
20 to 25 s : Lift decelerates at a constant rate of
25 to 30 s : Lift at rest or stationary.
30 to 35 s : Lift moves downward with a constant acceleration of
35 to 40 s : Lift moving downward with constant velocity
of
40 to 50 s : Lift decelerates at a constant rate of and comes to rest.
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Solution :
c. i.
05 10 15 20 25 30 35 t (s)
-4
-2
2
4
v (m s1)
40 45 50
A1
A2 A3
A4 A5
v-t ofgraph under the area distance Total 54321 AAAAA