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Equations of motion
Higher Physics
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• Experiments show that at a particular place all bodies falling freely under gravity, in a vacuum or where air resistance is negligible, have the same constant acceleration irrespective of their masses.
• This acceleration towards the surface of the Earth, known as the acceleration due to gravity, is donated by g.
• Its magnitude varies slightly from place to place on the Earth´s surface and is approximately 9.8ms-2
Acceleration Due to Gravity
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The Effects of Air Resistance
• Air resistance depends on 2 things– Surface area– Velocity
• Air resistance increases as surface area increases
• Air resistance increases as the velocity increases
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Terminal Velocity
• As an object falls through the air, it accelerates, due to the force of attraction of the Earth. This force does not change.
• As the velocity increases, the air resistance, the force opposing the motion, increases, therefore the acceleration decreases.
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• If the object falls for long enough, then the air resistance (a force acting upwards) will equal the force of attraction of the Earth (the weight) (a force acting downwards)
• Now there are no net forces acting on the object (since the two forces balance) so it no longer accelerates, but travels at a constant velocity called its terminal velocity.
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• Terminal velocity depends on– The size
– Shape
– And weight of the object
• A sky diver has a terminal velocity of more than 50ms-1 (how many km per hour?)
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Equations of motion
• Linear motion symbols:
• u - initial velocity (m/s)
• v - final velocity (m/s)
• a - acceleration (m/s2)
• s - displacement (m)
• t - time taken (s)
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Equations of motion
• There are three equations of motion:
atuv 2
2
1atuts
asuv 222
•The equations of motion can be used for projectiles and objects moving in a straight line (cars etc).
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Equations of motion
• Example 1:A car is travelling with a velocity of 5 m/s. It then accelerates uniformly and travels a distance of 50m. The velocity reached is 15 m/s.
• (a) Find the acceleration of the car.• (b) Calculate the time taken to travel this
distance.
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Solution
• 1. Write down what you know:
• 2. S= 50m, u = 5m/s , v=15m/s , a=a
• 3. As you are trying to find ‘a’ just write the letter ‘a’
• 4. Select the equation to use.
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Solution
2
22
22
/2100
200
10025225
502515
2
sma
a
a
asuv
•(a) Find the acceleration of the car.
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Solution
st
t
t
atuv
5
102
2515
(b) Calculate the time taken to travel this distance.
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Projectiles
• A projectile is an object projected by force and continuing in motion by its own inertia. (It is not moving under the effects of an engine)
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Projectile motion
• A projectile’s motion can be simplified by considering the horizontal and vertical components separately. These are independent of each other.
• The subscript v is used for vertical motion and h is used for horizontal motion.
• A projectile follows the path shown below.
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• The horizontal component of velocity is constant throughout the projectiles motion.
Path of a Projectile
A
B
CD
E
• Thus Vh = Uh
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Video - Constant Horizontal Motion
• Watch the video. How does this show constant horizontal velocity?
• How do the initial horizontal velocities of the plane and bomb compare?
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Path of projectile
• The vertical component of the velocity changes throughout due to gravity.• A - Vertical velocity is maximum.• B - Vertical velocity decreases due to effect of gravity. • C - Maximum height, the vertical velocity is zero.
A
B
CD
E
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Path of projectile
A
B
CD
E
•D - the projectiles velocity then starts to increase due to the acceleration of gravity. The velocity will now be negative.
•E - the projectile lands with the same velocity that it started with but is negative in magnitude.
•For projectile problems assume that upwards is positive and downwards is negative.
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Dealing with projectile problems
• Use this format to record data available.Horizontal Verticaluh =t =sh =
uv=vv=av= -9.8 ms-2
sv=t =
tus hh Use equations of motion
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Example 2
• A golfer hits a ball with a velocity of 48ms-1 at an angle of 300 as shown below.
• (a) What is the horizontal and vertical components of velocity?.
• (b) What is the resultant velocity of the ball after 2 seconds.
48 ms-1
300
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Solution
• (a) Update diagram
• uh = 48 cos 30 = 41.56 ms-1
• uv = 48 sin 30 = 24 ms-1
48 ms-1
300
uh
uv
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• (b) Horizontal component is constant , so final velocity (vh) = 41.56ms-1
• Vertical velocity has changed , use format and equations of motion to find vv .
uv= 24ms-1
vv= ???av= -9.8 ms-2
sv=t = 2s
14.4
)28.9(24
msv
v
atuv
v
v
vv
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• (b) To complete question , calculate the resultant velocity.
• Use pythagoras to find resultant velocity
?? ms-1
vh= 41.56ms-1
vv= 4.4 ms-1
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Example 3
• A stone is dropped down a water well and it hits the surface of the water after 4s.
• What was the velocity of the stone when it hit the water?uv= 0ms-1
vv= ???av= -9.8 ms-2
sv=t = 4s
12.39
)48.9(0
msv
v
atuv
v
v
vv
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Example 4
• An object is thrown horizontally from a cliff and takes 6s to land. If it lands 60m away from the base of the cliff, find:
• (a) its horizontal velocity
• (b) the height of the cliff
• (c) its velocity (size and direction) on impact.
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Example 5
• A balloon is climbing vertically with a velocity of 5ms-1 when a sandbag is dropped from it and hits the ground 3 seconds later.
• Find• (a) The velocity of the sandbag as it hits the
ground.• (b) The original height of the sandbag before
it is dropped.