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Distance against time graphs

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Presentation on theme: "Distance against time graphs"— Presentation transcript:

1 Distance against time graphs

2 Constant speed distance The gradient of this graph gives the speed
time

3 Getting faster (accelerating)
distance time

4 A car accelerating from stop and then hitting a wall
distance time

5 Speed against time graphs

6 No movement speed time

7 Constant speed speed time

8 Getting faster? (accelerating)
speed Constant acceleration time

9 Getting faster? (accelerating)
v The gradient of this graph gives the acceleration speed a = v – u t (v= final speed, u = initial speed) u time

10 Getting faster? (accelerating)
speed The area under the graph gives the distance travelled time

11 A dog falling from a tall building (no air resistance)
speed Area = height of building time

12 Acceleration/time graphs

13 Constant/uniform acceleration?
time

14 Note! The area under an acceleration/time graph gives the change in velocity acceleration time

15 Displacement Displacement the distance moved in a stated direction (the distance and direction from the starting point). A VECTOR

16 Displacement/time graphs
Usually in 1 dimension (+ = forward and - = backwards) Displacement/m Time/s

17 Velocity? Velocity is the rate of change of displacement. Also a VECTOR

18 Velocity/time graphs Usually in 1 dimension (+ = forward and - = backwards) Ball being thrown into the air, gradient = constant = m.s-2 velocity/m.s-1 Time/s

19 Acceleration? Acceleration is the rate of change of velocity. Also a VECTOR

20 Acceleration/time graphs
Usually in 1 dimension (+ = up and - = down) Acceleration = constant = m.s-2 accel/m.s-2 Time/s

21 Average speed/velocity?
Average speed/velocity is change in distance/displacement divided by time taken over a period of time.

22 Instantaneous speed/velocity?
Instantaneous speed/velocity is the change in distance/displacement divided by time at one particular time.

23 The equations of motion
The equations of motion can be used when an object is accelerating at a steady rate There are four equations relating five quantities u initial velocity, v final velocity, s displacement, a acceleration, t time SUVAT equations

24 The four equations 1 This is a re-arrangement of 2
NOT in data book The four equations 1 This is a re-arrangement of 2 This says displacement = average velocity x time 3 With zero acceleration, this becomes displacement = velocity x time 4 Useful when you don’t know the time

25 Beware! All quantities are vectors (except time!). These equations are normally done in one dimension, so a negative result means displacement/velocity/acceleration in the opposite direction.

26 Example 1 Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest?

27 Example 1 Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest? What does the question tell us. Write it out.

28 Example 1 Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest? u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = ? s

29 Example 1 Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest? u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = ? s Choose the equation that has these quantities in v = u + at

30 Example 1 Mr Blanchard is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his decceleration is 2 ms-2 how long will it take for the car to come to rest? u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = ? s v = u + at 0 = t 2t = 10 t = 5 seconds

31 Example 2 Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan?

32 Example 2 Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan? What does the question tell us. Write it out.

33 Example 2 Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan? u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = 5 s s = ? m

34 Example 2 Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan? u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = 5 s s = ? m Choose the equation that has these quantities in v2 = u2 + 2as

35 Example 2 Jan steps into the road, 30 metres from where Mr Blanchard’s engine stops working. Mr Blanchard does not see Jan. Will the car stop in time to miss hitting Jan? u = 10 ms-1 v = 0 ms-1 a = -2 ms-2 t = 5 s s = ? m v2 = u2 + 2as 02 = x-2s 0 = s 4s = 100 s = 25m, the car does not hit Jan. 

36 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.

37 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
When is the velocity of the ball 12 m.s-1?

38 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
When is the velocity of the ball 12 m.s-1? u = 24 m.s-1 a = -9.8 m.s-2 v = 12 m.s-1 t = ?

39 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
When is the velocity of the ball 12 m.s-1? u = 24 m.s-1 a = -9.8 m.s-2 v = 12 m.s-1 t = ? v = u + at

40 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
When is the velocity of the ball 12 m.s-1? u = 24 m.s-1 a = -9.8 m.s-2 v = 12 m.s-1 v = u + at 12 = t -12 = -9.8t t = 12/9.8 = 1.2 seconds

41 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
When is the velocity of the ball -12 m.s-1?

42 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
When is the velocity of the ball -12 m.s-1? u = 24 m.s-1 a = -9.8 m.s-2 v = -12 m.s-1 t = ?

43 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
When is the velocity of the ball -12 m.s-1? u = 24 m.s-1 a = -9.8 m.s-2 v = -12 m.s-1 v = u + at

44 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
When is the velocity of the ball -12 m.s-1? u = 24 m.s-1 a = -9.8 m.s-2 v = -12 m.s-1 v = u + at -12 = t -36 = -9.8t t = 36/9.8 = 3.7 seconds

45 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the displacement of the ball at those times? (t = 1.2, 3.7)

46 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the displacement of the ball at those times? (t = 1.2, 3.7) t = 1.2, v = 12, a = -9.8, u = 24 s = ?

47 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the displacement of the ball at those times? (t = 1.2, 3.7) t = 1.2, v = 12, a = -9.8, u = 24 s = ? s = ut + ½at2 = 24x1.2 + ½x-9.8x1.22 s = 28.8 – = 21.7 m

48 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the displacement of the ball at those times? (t = 1.2, 3.7) t = 3.7, v = 12, a = -9.8, u = 24 s = ? s = ut + ½at2 = 24x3.7 + ½x-9.8x3.72 s = 88.8 – = 21.7 m (the same?!)

49 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the velocity of the ball 1.50 s after launch?

50 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the velocity of the ball 1.50 s after launch? u = 24, t = 1.50, a = -9.8, v = ?

51 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the velocity of the ball 1.50 s after launch? u = 24, t = 1.50, a = -9.8, v = ? v = u + at

52 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the velocity of the ball 1.50 s after launch? u = 24, t = 1.50, a = -9.8, v = ? v = u + at v = x1.50 = 9.3 m.s-1

53 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the maximum height reached by the ball?

54 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the maximum height reached by the ball? u = 24, a = -9.8, v = 0, s = ?

55 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the maximum height reached by the ball? u = 24, a = -9.8, v = 0, s = ? v2 = u2 + 2as 0 = x-9.8xs 0 = s

56 Example 3 A ball is thrown upwards with a velocity of 24 m.s-1.
What is the maximum height reached by the ball? u = 24, a = -9.8, v = 0, s = ? 0 = s 19.6s = 242 s = 242/19.6 = 12.3 m

57 Imagine a dog being thrown out of an aeroplane.
Woof! (help!)

58 Force of gravity means the dog accelerates
To start, the dog is falling slowly (it has not had time to speed up). There is really only one force acting on the dog, the force of gravity. The dog falls faster (accelerates) due to this force. gravity

59 Gravity is still bigger than air resistance
As the dog falls faster, another force becomes bigger – air resistance. The force of gravity on the dog of course stays the same The force of gravity is still bigger than the air resistance, so the dog continues to accelerate (get faster) gravity

60 Gravity = air resistance Terminal velocity
As the dog falls faster and air resistance increases, eventually the air resistance becomes as big as (equal to) the force of gravity. The dog stops getting faster (accelerating) and falls at constant velocity. This velocity is called the terminal velocity. Air resistance gravity

61 Falling without air resistance
This time there is only one force acting in the ball - gravity gravity

62 Falling without air resistance
The ball falls faster…. gravity

63 Falling without air resistance
The ball falls faster and faster……. gravity

64 Falling without air resistance
The ball falls faster and faster and faster……. It gets faster by 9.81 m/s every second (9.81 m/s2) This number is called “g”, the acceleration due to gravity. gravity

65 Falling without air resistance?
distance time

66 Falling without air resistance?
speed Gradient = acceleration = 9.8 m.s-2 time

67 Velocity/time graphs Taking upwards are the positive direction
Ball being thrown into the air, gradient = constant = m.s-2 velocity/m.s-1 Time/s

68 Falling with air resistance?
distance time

69 Falling with air resistance?
Terminal velocity velocity time

70 Gravity What is gravity?

71 Gravity Gravity is a force between ALL objects! Gravity

72 Gravity The size of the force depends on the mass of the objects. The bigger they are, the bigger the force! Small attractive force Bigger attractive force

73 Gravity The size of the force also depends on the distance between the objects.

74 Gravity We only really notice the gravitational attraction to big objects! Hola! ¿Como estas?

75 Gravity The force of gravity on something is called its weight. Because it is a force it is measured in Newtons. Weight

76 Gravity 800 N On the earth, Mr George’s weight is around 800 N.
I love physics! 800 N

77 Gravity On the moon, his weight is around 130 N. Why? 130 N

78 Gravity In deep space, far away from any planets or stars his weight is almost zero. (He is weightless). Why? Cool!

79 Mass Mass is a measure of the amount of material an object is made of and also its resistance to motion (inertia). It is measured in kilograms.

80 Mass Mr George has a mass of around 77 kg. This means he is made of 77 kg of blood, bones, hair and poo! 77kg

81 Mass On the moon, Mr George hasn’t changed (he’s still Mr George!). That means he still is made of 77 kg of blood, bones, hair and poo! 77kg

82 Gravity In deep space, Mr George still hasn’t changed (he’s still Mr George!). That means he still is made of 77 kg of blood, bones, hair and poo! I feel sick! 77kg

83 Weight (N) = mass (kg) x gravitational field strength (N/kg)
Calculating weight To calculate the weight of an object you multiply the object’s mass by the gravitational field strength wherever you are. Weight (N) = mass (kg) x gravitational field strength (N/kg)

84 Newton’s 1st Law An object continues in uniform motion in a straight line or at rest unless a resultant external force acts

85 Newton’s first law Galileo imagined a marble rolling in a very smooth (i.e. no friction) bowl.

86 Newton’s first law If you let go of the ball, it always rolls up the opposite side until it reaches its original height (this actually comes from the conservation of energy).

87 Newton’s first law No matter how long the bowl, this always happens.
constant velocity

88 Newton’s first law Galileo imagined an infinitely long bowl where the ball never reaches the other side!

89 Newton’s first law The ball travels with constant velocity until its reaches the other side (which it never does!). Galileo realised that this was the natural state of objects when no (resultant ) forces act. constant velocity

90 Another example Imagine Mr George cycling at constant velocity.

91 Newton’s 1st law He is providing a pushing force. Constant velocity

92 Newton’s 1st law There is an equal and opposite friction force.
Pushing force friction Constant velocity

93 Newton’s second law Newton’s second law concerns examples where there is a resultant force.

94 Let’s go back to Mr George on his bike.
Remember when the forces are balanced (no resultant force) he travels at constant velocity. Pushing force friction Constant velocity

95 Newton’s 2nd law Now lets imagine what happens if he pedals faster.
Pushing force friction

96 Newton’s 2nd law His velocity changes (goes faster). He accelerates!
Remember from last year that acceleration is rate of change of velocity. In other words acceleration = (change in velocity)/time Pushing force friction acceleration

97 Newton’s 2nd law Now imagine what happens if he stops pedalling.
friction

98 Newton’s 2nd law So when there is a resultant force, an object accelerates (changes velocity) Mr George’s Porche Pushing force friction

99 It’s physics, there’s always a mathematical relationship!
Newton’s 2nd law There is a mathematical relationship between the resultant force and acceleration. Resultant force (N) = mass (kg) x acceleration (ms-2) It’s physics, there’s always a mathematical relationship! FR = ma

100 An example Resultant force = 100 – 60 = 40 N FR = ma 40 = 100a
a = 0.4 m/s2 Mass of Mr George and bike = 100 kg Pushing force (100 N) Friction (60 N)

101 Newton’s 3rd law If a body A exerts a force on body B, body B will exert an equal but opposite force on body A. Hand (body A) exerts force on table (body B) Table (body B) exerts force on hand (body A)

102 Free-body diagrams Shows the magnitude and direction of all forces acting on a single body The diagram shows the body only and the forces acting on it.

103 Examples Mass hanging on a rope T (tension in rope) W (weight)

104 Examples Inclined slope
If a body touches another body there is a force of reaction or contact force. The force is perpendicular to the body exerting the force Inclined slope R (normal reaction force) F (friction) W (weight)

105 Examples String over a pulley T (tension in rope) T (tension in rope)
W1 W1

106 Momentum Momentum is a useful quantity to consider when thinking about "unstoppability". It is also useful when considering collisions and explosions. It is defined as Momentum (kg.m.s-1) = Mass (kg) x Velocity (m.s-1) p = mv

107 Law of conservation of momentum
The law of conservation of linear momentum says that “in an isolated system, momentum remains constant”. We can use this to calculate what happens after a collision (and in fact during an “explosion”).

108 Law of conservation of momentum
In a collision between two objects, momentum is conserved (total momentum stays the same). i.e. Total momentum before the collision = Total momentum after Momentum is not energy!

109 A harder example! A car of mass 1000 kg travelling at 5 m.s-1 hits a stationary truck of mass 2000 kg. After the collision they stick together. What is their joint velocity after the collision?

110 A harder example! Before
2000kg 1000kg 5 m.s-1 Momentum before = 1000x x0 = 5000 kg.m.s-1 Combined mass = 3000 kg After V m.s-1 Momentum after = 3000v

111 Momentum before = momentum after
A harder example The law of conservation of momentum tells us that momentum before equals momentum after, so Momentum before = momentum after 5000 = 3000v V = 5000/3000 = 1.67 m.s-1

112 Momentum is a vector Momentum is a vector, so if velocities are in opposite directions we must take this into account in our calculations

113 An even harder example! Snoopy (mass 10kg) running at 4.5 m.s-1 jumps onto a skateboard of mass 4 kg travelling in the opposite direction at 7 m.s-1. What is the velocity of Snoopy and skateboard after Snoopy has jumped on? I love physics

114 An even harder example! Because they are in opposite directions, we make one velocity negative 10kg -4.5 m.s-1 4kg 7 m.s-1 Momentum before = 10 x x 7 = = -17 14kg v m.s-1 Momentum after = 14v

115 Momentum before = Momentum after
An even harder example! Momentum before = Momentum after -17 = 14v V = -17/14 = m.s-1 The negative sign tells us that the velocity is from left to right (we choose this as our “negative direction”)

116 “Explosions” - recoil

117 Impulse = Change in momentum
Ft = mv – mu The quantity Ft is called the impulse, and of course mv – mu is the change in momentum (v = final velocity and u = initial velocity) Impulse = Change in momentum

118 Impulse Ft = mv – mu F = Δp/Δt

119 Units Impulse is measured in N.s (Ft) or kg.m.s-1 (mv – mu)

120 In this case mv – mu = -3m -5m = -8m
5 m/s -3 m/s Impulse Note; For a ball bouncing off a wall, don’t forget the initial and final velocity are in different directions, so you will have to make one of them negative. In this case mv – mu = -3m -5m = -8m

121 Example Szymon punches Eerik in the face. If Eerik’s head (mass 10 kg) was initially at rest and moves away from Szymon’s fist at 3 m/s, what impulse was delivered to Eerik’s head? If the fist was in contact with the face for 0.2 seconds, what was the force of the punch? m = 10kg, t = 0.2, u = 0, v = 3 Impulse = Ft = mv – mu = 10x3 – 10x0 = 30 Ns Impulse = Ft = 30 Fx0.2 = 30 F = 30/0.2 = 150 N

122 Another example Impulse = mv – mu = = 0.3x-6 – 0.3x3 = -2.7kg.m.s-1
A tennis ball (0.3 kg) hits a racquet at 3 m/s and rebounds in the opposite direction at 6 m/s. What impulse is given to the ball? Impulse = mv – mu = = 0.3x-6 – 0.3x3 = -2.7kg.m.s-1 3 m/s -6 m/s

123 Area under a force-time graph = impulse
Area = impulse

124 Work In physics, work has a special meaning, different to “normal” English.

125 Work In physics, work is the amount of energy transformed (changed) when a force moves (in the direction of the force)

126 Calculating work The amount of work done (measured in Joules) is equal to the force used (Newtons) multiplied by the distance the force has moved (metres). Force (N) Distance travelled (m)

127 Work (J)= Force(N) x distance(m)
W = Fscosθ

128 Important The force has to be in the direction of movement. Carrying the shopping home is not work in physics!

129 Work = Fscosθ What if the force is at an angle to the distance moved?

130 Lifting objects Our lifting force is equal to the weight of the object. Lifting force weight

131 Work done (J) = Force (N) x distance (m)
A woman pushes a car with a force of 400 N at an angle of 10° to the horizontal for a distance of 15m. How much work has she done? W = Fscosθ = 400x15x0.985 W = 5900 J

132 Work done (J) = Force (N) x distance (m)
A man lifts a mass of 120 kg to a height of 2.5m. How much work did he do? Force = weight = 1200N Work = F x d = 1200 x 2.5 Work = 3000 J

133 Power! Power is the amount of energy transformed (changed) per second. It is measured in Watts (1 Watt = 1 J/s) Power = Energy transformed time

134 Work done in stretching a spring
Work done in strectching spring = area under graph F/N x/m

135 Chemical kinetic gravitational
Gain in GPE = work done = m x g x Δh

136 ΔEp = mgΔh m kg Joules N/kg or m/s2

137 Example Woof! (help!) A dog of mass 12 kg falls from an aeroplane at a height of 3.4 km. How much gravitational energy does the dog lose as it falls to the ground

138 Example On earth g = 10 m/s2 Mass of dog = 12 kg
Height = 3.4 km = 3400 m

139 Example On earth g = 10 m/s2 Mass of dog = 12 kg
GPE lost by dog = mgh = 12 x 10 x 3400 = J Height = 3.4 km = 3400 m

140 Just before the dog hits the ground, what has this GPE turned into?
Example GPE lost by dog = mgh = 12 x 10 x 3400 = J Just before the dog hits the ground, what has this GPE turned into?

141 Kinetic energy Kinetic energy of an object can be found using the following formula Ek = mv2 2 where m = mass (in kg) and v = speed (in m/s)

142 Example A bullet of mass 150 g is travelling at 400 m/s. How much kinetic energy does it have? Ek = mv2/2 = (0.15 x (400)2)/2 = J

143 Energy changes

144 Energy transfer (change)
A lamp turns electrical energy into heat and light energy

145 Sankey Diagram A Sankey diagram helps to show how much light and heat energy is produced

146 Sankey Diagram The thickness of each arrow is drawn to scale to show the amount of energy

147 Sankey Diagram Notice that the total amount of energy before is equal to the total amount of energy after (conservation of energy)

148 Efficiency Although the total energy out is the same, not all of it is useful.

149 Efficiency Efficiency is defined as Efficiency = useful energy output
total energy input

150 Example Efficiency = 75 = 0.15=15% 500

151 Energy efficient light bulb
Efficiency = 75 = 0.75 100 That’s much better!

152 Elastic collisions No loss of kinetic energy (only collisions between subatomic particles)

153 Inelastic collisions Kinetic energy lost (but momentum stays the same!)

154 Satellites

155 How far could you kick a dog?
From a table, medium kick.

156 How far can you kick a dog?
Gravity

157 Harder kick Gravity

158 Small cannon Woof! (help) Gravity

159 Bigger cannon Gravity Gravity

160 Even bigger cannon Gravity Gravity Gravity

161 VERY big cannon Gravity

162 Humungous cannon?

163 Dog in orbit! The dog is now in orbit! (assuming no air resistance of course)

164 Dog in orbit! The dog is falling towards the earth, but never gets there!

165 Dogs in orbit! The force that keeps an object moving in a circle is called the centripetal force (here provided by gravity) Gravity

166 Other examples Earth’s gravitational attraction on moon

167 Uniform Circular Motion
This describes an object going around a circle at constant speed

168 Direction of centripetal acceleration/force
Change in velocity VB VA VB VA VA + change in velocity = VB The change in velocity (and thus the acceleration) is directed towards the centre of the circle.

169 Uniform circular motion
The centripetal acceleration/force is always directed towards the centre of the circle Centripetal force/acceleration velocity

170 Not uniform velocity Uniform speed ≠ uniform velocity
It is important to remember that though the speed is constant, the direction is changing all the time, so the velocity is changing. Uniform speed ≠ uniform velocity

171 How big is the centripetal acceleration?
a = v2 = 4π2r r T2 where a is the centripetal acceleration (m.s-2), r is the radius of the circle (m), and v is the constant speed (m.s-1).

172 How big is the centripetal force?
F = mv2 r from F = ma (Newton’s 2nd law) Centripetal Force - The Real Force

173 Work done? None! Because the force is always perpendicular to the motion, no work is done by the centripetal force.

174 That’s it!


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