l.o. work out acceleration in a straight line using v i , v f , d and t
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L.O.Explain the following terms (and use appropriate units): Distance, displacement, speed, velocity, average speed, average velocity & acceleration. L.O. Work out acceleration in a straight line using v i , v f , d and t. L.O.Interconvert between a, v i and v f using a, d and t. Do now - PowerPoint PPT PresentationTRANSCRIPT
L.O. Work out acceleration in a straight line using vi, vf, d and t.
L.O. Explain the following terms (and use appropriate units): Distance, displacement, speed, velocity, average speed, average velocity & acceleration.
1. What is the symbol for distance?2. What is the symbol for speed?3. What is the symbol for time?4. What is the unit for distance?5. What is the unit for time?
Do nowQuietly, answer the following questions in your book
dvtmetersseconds
L.O. Interconvert between a, vi and vf using a, d and t.
Physical QuantitiesQuantity Symbol Unit
average velocity v ms-1
initial velocity vi ms-1
final velocity vf ms-1
distance d m
time t s
acceleration a ms-2
Determining the velocity of an object.The average velocity of an object can be determined by measuring the distance traveled and timing how long it takes to travel the distance.
Distance = 10 m
START FINISH
time = 2 s
ΔtΔdv
2)v(v
v if
Work out the average velocity of the following.
1. A car traveling 100 m in 4.0 s.
2. A cycle moving 20 meters in 8.0 seconds.
3. A mouse steadily accelerates from 5.0ms-1 to 10 ms-1.
4. The space shuttle moving 480 km in one minute.
5. A car traveling 300 km in 4.0 hours.
tdv
1-ms254s
100mv
tdv 1-ms5.2
820
v
2)v(v
v if
2105 )(v
15.7 msv
tdvaverage
1-average ms8000
60480000v
d = 480 km = 480 000 m
t = 1 min = 60 s
d = 300 km = 300 000 m
t = 4 hr = 240 min = 14400 stdvaverage
1-1-average ms21ms83.20
14400s300000mv
L.O. Use kinematic equations to solve problems involving constant acceleration in a straight line. (vertical motion)
Acceleration
Acceleration is the rate of change of velocity. In other words it is a measure of how quickly an object changes speed. Acceleration is always caused by an unbalanced force. The symbol for acceleration is a and the unit for it is ms-2(Meters per second per second).An object that speeds up, slows down or changes direction has accelerated. When a object slows down we say it decelerates. Deceleration is negative acceleration.
We use the following equation to work out the acceleration of an object.
tva
Final velocity (vf).The final velocity of an object is an instantaneous velocity and we can work it out by adding the change in the velocity to the initial velocity (vi).
atvv if Change in velocity as ∆v = at.
We use this equation to work out the final velocity of an object that has undergone constant acceleration for a known time.e.g. What is the final velocity of a car that was traveling at 20 ms-1 and accelerated constantly at 4 ms-2 for 6 s?
1ms486420 fv
Distance (d) or height.The equation on the right is used to work out the distance an object is travelling when it is accelerating at a constant rate. Often the equation is used to work out the height an object has fallen using the time it took to fall and the acceleration due to gravity (g = 9.8 ms-2)
d = ?
A ball is dropped from the top of the Eiffel tower it takes 8.13 s to hit the ground. What if the height of the tower?
2
21 attvd i
2
21 attvd i
213.88.92113.800.0 d
As the ball is not moving before it is dropped its initial velocity is zero ms-1.
320mm324 d
The equation on the right is used to work out the final velocity of an object if only the distance and acceleration is known. Often the equation is used to work the velocity of falling object just before they hit the ground, remember is the case of falling acceleration is always 9.8 ms-2.
d = 324 m
A ball is dropped from the top of the Eiffel tower. What is the balls velocity just before it hits the ground?
advv if 222
advv if 222
3248.9200.0 22 fv4.63502 fv
-1ms80689.794.6350 fv
2)( tvv
d fi
The distance an object travels can be worked out by multiplying it’s average velocity by the time
Average velocity.
2)( tvv
d fi
0 2 4 6 8 10 12 140
5
10
15
20
25
30
35
40
Distance/time graph
Time (S)
dist
ance
(m)
A curve indicates the object is accelerating
Horizontal line indicates the object is stationary
A straight line indicates constant speed
L.O. Explain the motion taking place and obtain information from displacement/time and velocity/time graphs.
0 1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
12
Distance a boy walks
Time (s)
Dis
tanc
e (m
)
Change in time
Change in distance
12ms4s8m
timedistancespeed
Working out the Speed from a Distance/time graph
0 2 4 6 8 10 120
1
2
3
4
5
6
7
velocity/time graph
Time (s)
velo
city
(m/s
)Object is moving at a constant velocity.
Object is decelerating.
Object is accelerating
0 2 4 6 8 10 120
1
2
3
4
5
6
7
Velocity/time graph
Time (s)
Velo
city
(m/s
)
Area of A = ½ base x height= ½ 6 x 6 = 18 m
Area of B = base x height= 3 x 6 = 18 m
Area of C = ½ base x height= ½ 2 x 6 = 6 m
Total Area = 18m + 18 m + 6m= 42 m
Distance travelled = 42 m
The total area under a velocity/time graph is equal to the distance and object has travelled
BCA
0 2 4 6 8 10 120
1
2
3
4
5
6
7
Velocity of Skateboard
Time(s)
Velo
city
(ms-
1)Do Now1. Work out the acceleration for parts A and B on the
graph.2. Work out the total distance the skateboard travelled.
Answers
21
0.5ms4s
2msa
21
1ms4s
4msa
Distance =½ 4x2+ +½4x4+ 6x2=4+4+8+12=24
Motion due to GravityThe gravity of the earth can have a huge effect on the motion of objects. All objects at sea level on the surface of the earth have a uniform gravitational field of 9.8 NKg-1 acting on them. That is a force, of 9.8 Newtons for every Kg of their mass, pulls them to the centre of the Earth.When objects are unsupported this force causes objects to accelerate at 9.8 ms-2 towards the centre of the Earth.
Projectile motionAn object undergoes projectile motion when the only force acting on it is gravity. • The force of gravity (g) causes all objects to always accelerate
vertically towards the ground at 9.8 ms-2. • There are no force acing horizontally on an object, so there is no
acceleration and the object moves at a constant speed.
1. Describe the horizontal motion of this ball.
2. Describe the vertical motion of this ball.
vi = 20 ms-1
θ = 50°
vh = 13 ms-1
vi = 20 ms-1
The motion of an object moving through the air we can simplified by separating it into two components: Vertical motion and horizontal motion.
vv = 15 ms-1
vi = 20 ms-1
Manipulating formula
d
v t
tdv
tvd
vdt
Distance is the subject of the formula
Time is the subject of the formula
Speed is the subject of the formula
Work out the distance travelled by the following.
1. A car travelling for 100 seconds at 20 meters per seconds.
2. A motor cycle moving for 200 seconds at 6 meters per seconds.
3. A cat running 50 seconds in 10 metres per.
Work out the time taken to travel the following.
1. A car travelling 50 metres at 15 meters per seconds.
2. A boat moving 500 metres at 6 meters per seconds.
3. A dog running 20 metres in 4 metres per.
Speed
• Average speed is calculated for the complete journey.
• Instantaneous speed is the actual speed of any part of the journey.
• What does the speedometer in the car measure?
• Actual (instantaneous) speed.
Speed
• Average speed (Vav) is calculated by the formula:
• Vaverage = distance travelled• total time taken
• Vav = d• t
Units involved
Distance Time Speed
Centimetre (cm) Second (s) cms-1
Metre (m) Second (s) ms-1
Kilometre (km) Hour (hr) Kmh-1
Example
• John travels from Dunedin to Christchurch.• It is 325km and it takes him 4.5 hours.• What is his average speed?
tdvaverage
Answer the following questions.
1. What is the average speed of a car travelling 250 metres in 10 seconds?
2. How long does it take a boat to sale 500 metres at 12 meters per seconds?
3. How for will a snail moving at 0.05 ms-1 travel in 2 min?
•Draw distance/time or speed/time graphs from data•Use the gradient (slope) of a distance/time graph to describe the speed of an object
250/10 = 25 ms
500/12 = 42.5 s
0.05 x 120 = 6 m
L.O. Define acceleration and state its unit and symbol
Acceleration
Acceleration is the rate of change of velocity. In other words it is a measure of how quickly an object changes speed. Acceleration is always caused by an unbalanced force. The symbol for acceleration is a and the unit for it is ms-2(Meters per second per second).An object that speeds up slows down or changes direction has accelerated. When a object slows down we say it decelerates. Deceleration is negative acceleration.
We use the following equation to work out the acceleration of an object.
tva
Use the acceleration equation to answer the following questions.
What is the acceleration of a car that takes 7.4 seconds, from standing still, to reach 45 ms-1?Change in speed = 45 – 0 = 45 ms-1 Change in time = 7.4 s
21
1.6081.64.7
45
mss
msa
What is the acceleration of a girl that takes 4.3 seconds to go from 0 ms-1 to 5.6 ms-1?Change in speed = 5.6 – 0 = 5.6 ms-1 Change in time = 4.3 s
21
3.1302.13.4
6.3
mssmsa
ExerciseComplete exercises 1-7 on page 163 of the New Directions in Science text book..
Calculate the time taken to reach a velocity of 13.5 ms-1 is a car is acceleration at 3.2 ms-2.
smsms
avt 2.4218.4
2.35.13
2
1
•Can draw free body force diagrams to show the forces acting on an object.
•Can work out the net force acting on an object by combining the force vectors from a force diagram.
•Can break force vectors into their horizontal and vertical (or North and West ) components using soh cah toa.
•Can use the equation F= m x a to work out one component when the others are known.
•Can interconvert between mass and weight using 9.8 ms-2 as the acceleration due to gravity.
•Can describe two examples to illustrate each of Sir Isaac Newton’ three Laws of Motion.
Force, mass and acceleration.
Force A force is a push, pull or a twist to an object that causes it to accelerate. The symbol for force is F and the unit is the Newton, N. Force is a vector quantity and this is often depicted by an arrow.
The resultant force is the product of the vectors of all the forces acting on an object.
15 N
9.8 N Resultant forceResultant force
Vector triangleDiagram of ball being thrown
Law 3For every action force there is an equal and opposite reaction force. (the action force acts on an object and the reaction force acts on a different object)
Law 2When the resultant force acting on an object is non-zero the object will accelerate in the direction of the force.
Law 1If the resultant force acting on an object is zero then the object will have a constant velocity.
Newton’s laws of motion.
amF
4N
Support6N
2N
6NGravity
FrictionThrust
The net force is 2N to the left This means that the toy car will accelerate to the left.
Net force
2N
Unballanced forces
The size of the force needed to accelerate an object at a certain rate is dependent on two factors:
• The rate of acceleration acting on the object. A faster acceleration will need a larger force.
• The mass of the object the force is acting on. The larger the mass, the large the force need to accelerate it.
Force, mass and acceleration.
amF ExampleWhat force is needed to accelerate and 10 kg ball 3 ms-2 ?F=m x a = 10 x 3 = 30 N
All objects, at sea level, on the earth are accelerated by gravity (g) at a rate of 9.8 ms-2:
• The force that is produced by this acceleration is called weight and it is measured in Newtons (N).
• This force of gravity (weight) can be calculated by using the following equation. g = 9.8 ms-2
Force of gravity.
gmFg ExampleWhat is weight acting on a 60 kg girl at sea level?Fg=m x g = 60 x 9.8 = 588 N
W f d Work is done when an object is moved a distance (d) by and force (f) that is parallel to the movement. Only the component of the force that is parallel to the direction of the movement.
d = 12 m
f = 20 Nθ = 30˚
Power
EPt
Power is how much energy(work) is produced per second. When using this equation the energy could come from work, kinetic energy or Ep.
ExampleWhat is the power of a motor that does 100 J of work in 4 seconds?Energy = work = 100JP = E ⁄ t = 100J / 4 s = 25 W
Power measured in Watts(W)
Energy measured in Joules(J)
Time measured in seconds(s)
ExampleHow much energy is used when a 100 W light bulb is on for 30 s?E = P x t = 100 W x 30 sE = 3000 J
tPE
PEt
ExampleHow long does it take a 500 W motor to do 10 000 J work?t = E / P = 500 W / 10 000 jt = 20 s
Rearranging the EquationTo find the energy
To find the time
Kinetic energyKinetic energy is the energy that moving objects have. We use the following equation to work out the kinetic energy an object has.
2k mv
21E
ExerciseWhat is the kinetic energy of a 8.0 kg ball moving at 5.0 ms-1?
Ek = ½ mv2 = ½ 8.0 x 52
= 4.0 x 25 = 100 J
mass(Kg)
speed(ms-1)
kinetic energy (J)
2k mv
21E
In this equation speed is squared. This has the effect of increasing the energy by four times when the speed is doubled.
ExampleBall traveling at 4 ms-1
Ek = ½ mv2 = ½ 8 x 42
= 4 x 16 = 64 J
Ball traveling at 8 ms-1 (Double the speed)Ek = ½ mv2 = ½ 8 x 82
= 4 x 64 = 256 J (The kinetic energy is quadrupled)
Gravitational potential energyGravitational potential energy (GPE) is the energy an object above the ground has. We use the following equation to work out the gravitational potential energy an object has.
mghEp
ExerciseWhat is the GPE of a 2.0 kg ball 8.0 m above the ground?
Ep = 2.0 kg x 9.8 ms-2 x 8 m = 160 J
mass(Kg)
Force of gravity(9.8 NKg-1)
Gravitational potential energy (J)
Height of object(m)
Converting energy
Any time that an object is lifted the work done to lift the object is equal to the gravitational potential energy (Ep) gained by the object.
2 m
Ep = mgh = 0.5 x 10 x 2 = 10 J
Mass of toy car 0.5 Kg
W = f x d = 5 x 2 = 10 J
What is the gravitational potential energy gained when a box is lifted with a constant force of 50 N for 1.5 m?W = f x d = 50 x 1.5 = 75 JEp = work done = 75J
• When work is done accelerating an object by a force over a distance, this energy is transformed into kinetic energy(Ek). However, because of friction some of this work is turned into heat energy and only some of the work is transformed into kinetic energy.
• Any time that energy is transformed some energy is always lost as heat energy.
• The amount of heat energy lost can be reduced by decreasing the amount of friction acting on an object.
• We can work out the efficiency of an energy transformation using the following equation.
1%100
__
inenergyoutenergyEfficiency
When on object falls, the gravitational potential energy it had at the top of the fall is equal to the kinetic energy it has just before it hits the ground (if we ignore air friction).
4 m
At the topEp = mgh = 2 x 10 x 4 = 80 J
Just before hitting the groundEk = Ep
Ek = 80 J
•Describe Hooke’s law•Interconvert between the force, spring constant or extension of a spring using
kxF -=Hooke’s law states that “the distance a spring extends is proportional to the tension force acting on the spring.”
10 N
0.12 m
The diagram on the right shows what happens when a 10 N force is applied to an un-stretched spring. The tension force (F) is always in the opposite direction to the force extending the spring. For this spring the tension force, F = -10 N and the extension, x = 0.12 m. Hooke’s law can be used to work out the spring constant for this spring.
Hooke’s Law
212pE kx
The potential energy gained when a spring is compressed or extended can be calculated by using the equation on the right. The potential energy of a spring is dependent only on the extension of the spring and the spring constant.
ExercisesComplete the following exercises from page 150 of NCEA Level 1 Science book.Questions 1, 2, 3 and 4
Answers
212pE kx