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OUTCOMES
Derive and use equations of motion with constant acceleration
Sketch and use the graphs of displacement-time, velocity-time and acceleration-time for the motion of a body with constant acceleration
Solve problems on projectile motion without air resistance
Explain the effects of air resistance on the motion of bodies in air
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2.1 LINEAR MOTION
Derive and use equations of motion with constant acceleration Sketch and use the graphs of displacement-time, velocity-time and acceleration-time for the motion of a body with constant acceleration
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2.1 LINEAR MOTION
Linear Motion: The object is moving along a straight line
Linear Motion with constant velocity: ... at constant velocity
Linear Motion with constant acceleration: … at a constant change rate of velocity.
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MOTION WITH CONSTANT ACCELERATION
Derivation: Let u = initial velocity, v = final velocity
after a time, t, then, from the definition, The uniform acceleration a = (change of velocity)/(time taken) a = (v u) / t …(1) v = u + at …(2)
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MOTION WITH CONSTANT ACCELERATION
v = u + at …(2) Let s = displacement of the body in time t, then s = (average velocity) (time) s = [(u + v)/2] t s = ½ (u + v)t …(3) (2) (3): s = ½ [u + (u + at)]t s = ut + ½ at2 …(4)
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MOTION WITH CONSTANT ACCELERATION
v = u + at…(2) s = ½ (u + v)t …(3) From (2): v – u = at …(5) v + u = 2s/t …(6) (6) (5): (v + u)( v – u ) = (2s/t)(at) v2 – u2 = 2as v2 = u2 + 2as …(7)
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MOTION WITH CONSTANT ACCELERATION
Reminder: a = (v u) / t v = u + at s = ½ (u + v)t s = ut + ½ at2
v2 = u2 + 2as
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EXAMPLE 1
1. An object which is initially at rest at the centre of coordinate 0 (x = 0) starts to move along the x-axis with constant acceleration. It moves to point P and then to point Q at velocity 15 m s1 and 25 m s1 respectively. The distance PQ = 100 m. Determine
(a) the acceleration of the object, (b) the distance OP.
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ANSWER 1
(a) Use v2 = u2 + 2as for motion between P and Q, 252 = 152 + 2a(s2 – s1); 252 = 152 + 2a(100)
a = 2 m s-2
(b) Use v2 = u2 + 2as for motion between O and P, 152 = 0 + 2(2) s1
s1 = 56.3 m
O Q S2 = S1 +100
U - 0 S1
V1 = 15 ms-1 V2 = 25 ms-1
100 m P
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EXAMPLE 2
2. An object moves along a straight line with constant acceleration. Its initial velocity is 20 m s1. After 5.0 s, the velocity becomes 40 m s1. Determine the distance travelled during the third second.
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ANSWER 2
Use v = u + at 40 = 20 + a(5) a = 4 m s-2
Use s = ut + ½ at2
U = 20 m s-1 S1 S2 v = 40 m s-1
t1 = 2 s
t2 = 3 s
Third second
s1 = 20t1 + ½ (4)t12
= 20t1 + 2t12 …(1)
s2 = 20t2 + ½ (4)t22
= 20t2 + 2t22 …(2)
(2) – (1): s2 -s1
= 20(t2 – t1) + 2(t22 – t1
2) = 30 m
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EXAMPLE 3
3. An object moves along a straight line with constant acceleration. At a particular moment, the velocity of the object is 30 m s-1. After travelling through 50 m, the object has velocity 20 m s1. How far does the object have to travel before it comes to a stop?
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ANSWER 3
Motion from P to Q: Use v2 = u2 + 2as 202 = 302 + 2a(50) a = -5 m s-2
P Q R
U = 30 ms-1 v = 20 ms-1
50 m s
Motion from Q to R:
Use v2 = u2 + 2as
0 = 202 + 2(-5)s
s = 40 m
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EXAMPLE 4
4. A car which is initially at rest starts to move along a straight line with constant acceleration. It reaches a velocity of 60 m s1 after travelling through a distance of 100 m. Determine
(a) the acceleration, (b) the time taken to reach the velocity
of 60 m s1, (c) the velocity at the third second.
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ANSWER 4
(a) Use v2 = u2 + 2as 602 = 0 + 2a(100) a = 18 m s-2
(b) Use v = u + at 60 = 0 + 18t t = 3.3 s (c) Use v = u + at
= 0 + (18) (3) = 54 m s-1
DISPLACEMENT-TIME GRAPH (S-T)
Note that the instantaneous velocity, v = ds/dt = gradient of the graph at time t. The instantaneous velocity at time 4s, v = (18-5)/(4-0) = 13/4 = 3.25 ms-1.
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S / m
t / s 0 2 4 6 8
10
20
5
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DISPLACEMENT-TIME GRAPH AZ (S-T)
At time 0s to 1s, the velocity is constant; from time 1s to 3s, the velocity is increasing and from time 3s to 5s, the velocity is decreasing.
S / m
t / s 0 2 4 6 8
10
20
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DISPLACEMENT-TIME GRAPH (S-T)
Negative gradient at time 6s onward shows that the object is moving in the opposite direction.
S / m
t / s 0 2 4 6 8
10
20
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VELOCITY-TIME GRAPH (V-T)
Note that the acceleration, a = dv/dt = gradient of graph. The acceleration at time 0.5 s is given by a = (20-0)/(2-0) = 10 ms-2
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v / ms-1
t / s 0 2 4 6 8
10
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VELOCITY-TIME GRAPH (V-T)
Time Acceleration
0 s to 1 s constant
3s to 5s decreasing
5s to 7s 0 ms-2 (Constant)
7s to 8s deceleration
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v / ms-1
t / s 0 2 4 6 8
10
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VELOCITY-TIME GRAPH (V-T)
The displacement within a period can be found by finding the area under the graph. Displacement between t = 0 and t = 1, s1 = ½ 1 20 = 10 m Displacement between t = 1 and t = 4, s2 = ∫1
4 vdt
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v / ms-1
t / s 0 2 4 6 8
10
20
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EXAMPLE 5
A body moves along the x-axis. Assume that a positive sign represents a direction to the right. The velocity v of the body is related to time t through the equation
v =2 3t2
where v and t are measured in m s-1 and s respectively. t = 0 when x = 0. Determine
(a) the displacement, (b) the acceleration, at the instant of time t = 1 s.
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ANSWER 5
(a) v = ds/dt = 2 – 3t2 s = ∫0
1(2 – 3t2)dt = [2t – t3]01 = 1.0 m
(b) a = dv/dt = -6t = -6(1) = -6 m s-2
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ACCELERATION-TIME GRAPH (A-T)
The change in velocity within a period can be found by finding the area under the graph. Change in velocity between t = 2 and t = 6, v = 15 (6 – 2) = 60 ms-1
OR, v = ∫26 a dt = ∫2
6 15 dt
= [15a]26 = 15 (6 -2) =
60ms-1
a / ms-2
t / s 0 2 4 6 8
10
20
Constant acceleration
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2.2 PROJECTILE (PROJECTION)
A projectile is a body (object) which is projected(or thrown) with some initial velocity, and then allowed to be acted upon by the forces of gravity and possible drag.
E.g. Baseball being thrown, water fountains, fireworks displays, soccer ball being kicked & ballistics testing.
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2.2 PROJECTILE (PROJECTION)
Definition: A projectile is an object
that is projected at an angle to the horizontal and moves under the action of gravity.
The path of a projectile is called its trajectory.
vo
h
x
projectile
trajectory
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2.2 PROJECTILE (PROJECTION)
Height, h is the maximum upward distance reached by the projectil
Range, x is horizontal distance travelled (or sometimes distance).
vo
h
x
projectile
trajectory
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2.2 PROJECTILE (PROJECTION)
If a body is allowed to free-fall under gravity and is acted upon by the drag of air resistance, it reaches a maximum downward velocity known as the terminal velocity.
vo
h
x
projectile
trajectory
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2.2 PROJECTILE (PROJECTION)
The study of the motion of projectiles is called ballistics.
vo
h
x
projectile
trajectory
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2.2 PROJECTILE (PROJECTION)
An object falling freely towards the ground is exerted by gravitational force, g = +9.8ms-2. For object falling freely without
resistant, we have v = u + gt s = ut + ½ gt2
v2 = u2 + 2gs
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2.2 PROJECTILE (PROJECTION)
For an object moves upwards against the gravity, g = 9.8ms-2.
In this case, the displacement and velocity that pointing upwards have positive magnitude, and the magnitude is negative while pointing downwards.
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2.2 PROJECTILE (PROJECTION)
In general, an object falling freely to ground experience inconsistence air resistance, Fa.
The net force on the object, F = weight of object air
resistance
F = mg Fa
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2.2 PROJECTILE (PROJECTION)
The magnitude of Fa is not consistence but increases with the speed of the object.
When Fa = mg, the net force, F = 0 and therefore the velocity of
the object becomes constant.
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2.2 PROJECTILE (PROJECTION)
Since the acceleration of the object is given by a = g - Fa/m, the acceleration becomes zero when Fa increases with time.
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2.2 PROJECTILE (PROJECTION)
Projectile motion is a combination of uniform motion along x and uniformly accelerated motion (free fall) along y.
= Sum of 2 independent motions Ignoring air resistance, horizontal motion, x
has constant velocity, while vertical motion, y is accelerated by gravitational force.
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2.2 PROJECTILE (PROJECTION)
Along x, the projectile travels with constant velocity, where vx= vxo and x = vxot
Vertical component, vy
Horizontal component, vx
Net velocity, v
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2.2 PROJECTILE (PROJECTION)
Along y, the projectile travels in free-fall fashion, where vy = vyo – gt, y = vyot – (1/2) gt2 and g = 9.8 m/s2
Vertical component, vy
Horizontal component, vx
Net velocity, v
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SUMMARY OF PROJECTILE MOTION
Motion Horizontal Vertical
Initial Velocity
u cos u sin
Acceleration 0 -g
Time taken t t
Distance moved
x = ut cos y = ut sin - gt2/2
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SUMMARY OF PROJECTILE MOTION
vy = u sin
vx = u cos
v = vx + vy
Greatest height, h = (u sin )2 / 2g at t = (u sin )/g Time of flight, T = (2 u sin ) / g
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SUMMARY OF PROJECTILE MOTION
Range, R = (u2 sin 2 ) / g Maximum horizontal range, Rmax = u2 / g When = 450
vy = u cos
vx = u sin
v = vx + vy
Trajectory, y = x tan - gx2/(2u2 cos2 ) Or y2 = 2x(ucos )2/g
2.3 FREE FALLS AND AIR RESISTANCE
Explain the effects of air resistance on the motion of bodies in air
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AIR RESISTANCE
• In air… – A stone falls faster
than a feather • Air resistance affects
stone less • In a vacuum
– A stone and a feather will fall at the same speed
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AIR RESISTANCE
• Effects of air resistance – A person in free fall
reaches a terminal velocity of around 54 m/s
– With a parachute, terminal velocity is only 6.3 m/s
• Allows a safe landing
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GALILEO GALILEI 1564 - 1642 • Applied scientific method • formulated the laws that
govern the motion of objects in free fall
• Also looked at: • Inclined planes • Relative motion • Thermometers • Pendulum
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FREE FALL
All objects moving under the influence of gravity only are said to be in free fall Free fall does not depend on the object’s
original motion All objects falling near the earth’s surface
fall with a constant acceleration due to gravity
Acceleration due to gravity is independent of mass
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ACCELERATION DUE TO GRAVITY
Symbolized by g g = 9.80 m/s² When estimating, use g » 10 m/s2
g is always directed downward Toward the center of the earth
Ignoring air resistance and assuming g doesn’t vary with altitude over short vertical distances, free fall is constantly accelerated motion
FREE FALL
Definition: motion when gravity is the only force acting on an object
This is hard to achieve in the classroom, laboratory, and even outdoors because of Air Resistance.
For our working definition, assume little to no Air Resistance.
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FREE FALL – AN OBJECT DROPPED
Initial velocity is zero Up is generally taken to
be positive In the kinematic
equations, generally use y instead of x
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vo= 0
a = g
FREE FALL – AN OBJECT THROWN DOWNWARD
Free upon release With up being positive,
initial velocity will be negative
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vo= 0
a = g
FREE FALL - OBJECT THROWN UPWARD
Initial velocity is upward, so positive
The instantaneous velocity at the maximum height is zero
a = g = -9.80 m/s2 everywhere in the motion
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v = 0
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THROWN UPWARD, CONT.
The motion may be symmetrical Then tup = tdown
Then v = -vo
The motion may not be symmetrical Break the motion into various parts
Generally up and down
NON-SYMMETRICAL FREE FALL
Need to divide the motion into segments
Possibilities include Upward and downward
portions The symmetrical portion
back to the release point and then the non-symmetrical portion
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COMBINATION MOTIONS
Example: Launching of rocket
Phase 1: a = 29.4ms-2 (fuel burns out)
Phase 2: a = -9.80 ms-2 (rocket falls freely after reaches maximum height)
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EXAMPLE 6: FREE FALL
A ball is dropped from rest from the top of a building. Find: a) The instantaneous velocity of the ball
after 6 sec. b) How far the ball fell. c) The average velocity up to that point. Answers: -60m/s, 180m, -30m/s
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EXAMPLE 7: FREE FALL ON THE MOON
A hammer is dropped on the moon. It reaches the ground 1s later. If the distance it fell was 0.83m: a) Calculate the acceleration due to gravity on
the surface of the moon. b) Calculate the velocity with which the hammer
reached the ground and compare to the velocity it would have, if it was dropped on the earth’s surface.
Answer: -1.66m/s2, -1.66m/s, -9.8m/s
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