Kinematics
 What is gravity?
 Where does it come from?
 What kinds of things have gravity?
 I have two masses. One is a basketball. The other is a
5lb bag of potatoes. If I hold them at the same height
and drop them at the same time, which one hits the
ground first?
 Do they have a constant velocity?
 Do they have the same gravitational force acting on
them?
Equations!!!
 Equations in IB look different.
 Fill in the following chart based on the given
equations.
Concept
Variable
Unit
Vocabulary
 Distance – a scalar quantity of position changed
 Displacement - a vector quantity of a change in the
position of an object.
 Speed – is the rate of change of distance with respect to
time. (scalar)
 Velocity – is the rate of change of displacement with
respect to time. (vector quantity)
 Acceleration – is the rate of change of velocity with
respect to time. (vector quantity)
Distance and Displacement
 A mass initially at 0m moves 10m to the right and then
2m to the left. What is the final displacement? What is
the final distance traveled?
 A mass initially at 0m, first moves 5m to the right and
then 12m to the left. What is the total distance covered
by the mass and what is the change in displacement?
Distance and Displacement
 Distance = 12m, Displacement = 8m
 Distance = 17m, Displacement = -7m
Speed and Velocity
 Speed is nornally given in m/s or km/h.
 OR
ms-1 or kmh-1
 What the difference?
 Speed is a scalar, velocity is a vector.
Speed and Velocity
 A car of length 4.2m travelling in a straight line takes
0.56s to go past a mark on the road. What is the speed
of the car?
 A car starts out from the 100km mark in a straight line
and moves a distance of 20km towards the right, and
then returns to its starting position 1h later. What is
the average speed and the average velocity for this trip?
 V = 7.5 m/s
 Avg speed = 40km/h, avg velocity = 0km/h
 Starting point is irrelevant.
Equations
Acceleration
 We mostly look at constant acceleration situations.
 In this case the instantaneous acceleration and average
acceleration are the same thing.
 Acceleration due to gravity is 9.8m/s2 or 10m/s2 or
9.81m/s2.
 Don’t forget about the ball thrown upward.
 Don’t forget about the variable table!
Acceleration
 An object starting with an initial velocity of 2 m/s
undergoes constant acceleration. After 5s its velocity is
found to be 12m/s. What is the acceleration?
 A ball is thrown downward from a 70m tower with an
initial velocity of 3m/s. How fast would it be going
after 2 seconds? What would it’s position be after
2seconds?
Acceleration
 2m/s2
 22.6m/s, 44.4m high
Everything
 A mass has an initial velocity of 10m/s. It moves with
acceleration -2m/s2. When will it have a zero velocity?
 What is the displacement after 10s of a mass whose
initial velocity is 2m/s and moves with acceleration of
4m/s2?
 A car has an initial velocity of 5m/s. When its
displacement increases by 20m, its velocity becomes
7m/s. What is the acceleration?
 A body has initial velocity of 4m/s and a velocity of
12m/s after 6s. What is the displacement?
 t = 5s
 s = 220m
 a = .6m/s2
 s = 48m
Test yourself
 Two balls start out moving to the right with constant
velocities of 5m/s and 4m/s. The slow ball stars first
and the other 4s later. How far from the starting
position are they when they meet?
 A mass is thrown upwards with an initial velocity of
30m/s. A second mass is dropped from directly above,
a height of 60m from the first mass, 0.5s later. When
do the masses meet and how high is the point where
they meet?
 t=20s
so s = 80m
 t=2.35s so s is 42.9m
Graphing
Cheat Sheet
 The slope of d-t graph gives the value of the v-t graph
 The slope of v-t graph gives the value of the a-t graph
 Area under a-t graph gives the change in velocity
 Area under v-t graph gives the change in displacement.
Example
Bell Ringer
Sept 4/5
 Consider the following graph
Bell Ringer
Sept 4/5
 What is the initial displacement?
 What is the velocity for the first 10s? Second 5s?
 When is the object at the origin?
 What is Δs? What is the total distance traveled?
 What is the avg velocity?
 Initial displacement is -10m
 2m/s, -2m/s
 5s and 15s
 Δs = 10m, total distance = 30m
 Avg speed = 2m/s, avg velocity = .66m/s
Interpreting graphs
 A mass starts out from zero with velocity 10m/s and
continues moving at this velocity for 5s. The velocity is
then abruptly reversed to -5m/s and the object moves
at this velocity for 10s. For this event find:
a) The graph
b) The change in displacement
c) The total distance travelled
d) The average speed
e) The average velocity
a) See previous slide
b) 10 x 5m = 50m, -5 x 10 = -50m, Δs = 0
The object moved toward the right, stopped and returned
to its starting position.
c) 50m to the right, 50m to the left, 100m total
d) Avg velocity =0
e) Avg speed= 100m/15s = 6.7m/s
 Use the graph and your understanding of slope
calculations to determine the acceleration of the
rocket during the listed time intervals.
 t = 0 - 1 s, t = 1 - 4 s, t = 4 - 12 s
 a = 40 m s-2
 b = 20 m s-2
 c = - 20 m s-2
 Describe the motion depicted by the following velocity-time
graphs. In your descriptions, make reference to the direction
of motion (+ or - direction), the velocity and acceleration and
any changes in speed (speeding up or slowing down) during
the various time intervals (e.g., intervals A, B, and C).
 The object moves in the + direction at a constant speed - zero
acceleration (interval A). The object then continues in the +
direction while slowing down with a negative acceleration
(interval B). Finally, the object moves at a constant speed in
the + direction, slower than before (interval C).
 The object moves in the + direction while slowing down; this
involves a negative acceleration (interval A). It then remains
at rest (interval B). The object then moves in the - direction
while speeding up; this also involves a negative acceleration
(interval C).
 The object moves in the + direction with a constant velocity
and zero acceleration (interval A). The object then slows
down while moving in the + direction (i.e., it has a negative
acceleration) until it finally reaches a 0 velocity (stops)
(interval B). Then the object moves in the - direction while
speeding up; this corresponds to a - acceleration (interval C)
 Consider the velocity-time graph below. Determine the
acceleration (i.e., slope) of the object as portrayed by
the graph.
 The acceleration (i.e., slope) is 4 m/s/s. If you think
the slope is 5 m/s/s, then you're making a common
mistake. You are picking one point (probably 5 s, 25
m/s) and dividing y/x. Instead you must pick two
points (as in the discussed in this part of the lesson)
and divide the change in y by the change in x.
 Determine the displacement of the object during the
time interval denoted by the shaded area.
 a = 90m
 b = 45m
 c = 40m
Relative velocities
 http://regentsprep.org/Regents/physics/phys01/velocit
y/relative.htm
 A car (A) moves to the left with speed 40km/h (with
respect to the road). Another car (B) moves to the right
with speed 60km/h(also with respect to the road).
Find the relative velocity of B with respect to A.
 Do pg 46 #4,5, 6, 8, 9, 11, 12
FORCES!!!
 Types of forces
 Gravitational – between objects as a result of their masses. (also
called weight)
 Normal reaction – between two surfaces that are touching
 Frictional- force that opposes the relative motion of two
surfaces. (Includes Air resistance/drag)
 Applied – force of an outside object pushing or pulling
 Tension- when a string/spring is stretched, it is equal and
opposite the force acting on it.
 Compression – opposite of tension (applies to solid objects)
 Upthrust – the upward force that acts on objects when it is
submerged in a fluid. Causes objects to float.
 Lift – caused when air flows over an aircraft’s wing. Causes an
upward force.
Free Body Diagram
 Five Steps.
1) Is there gravitational force? (Fg)
2) Is it sitting on a surface? (Fn)
3) Is there some thing pushing or pulling? (Fapp)
4) Is there friction? (Ff )
5) Is it accelerating? a = ?
http://www.youtube.com/watch?v=BuPfDI7TyL0
 Forces are vectors
 Magnitude and direction
 Many times the direction is the x or y
 This means they can be resolved into components.
 This also means they can be added and subtracted.
http://www.youtube.com/watch?v=IrY-FlJ0c7Y

Practice/Application
A person is trying to lift a heavy concrete block
without success. The upward force exerted on the
block by the person is P, the contact force on the
block by the floor is C, the weight of the block is W.
 Which one of the following is true about the
magnitudes of the forces on the block while the person
is trying to lift it?
a) P + C =W
b) P + C < W
c) P+ C > W
d) P = C = W
1.
Practice/Application
2. Consider the vector with a magnitude of 7N acting at
an angle of 140° to the horizontal. What will the
horizontal and vertical components of this vector be?
3. An object O is acted upon by three forces as shown in
the diagram. What is the magnitude of the resultant
force acting on O?
Answers
1) Since the person is not succeeding to lift the box, the
weight of the box must be equal to the sum of the
upward force and the floor contact force. Answer: A
2) y= 5.4 x = -4.5
3) The 6N and 3N forces are acting against each other.
This results in a 3N force to the left. This resulting 3N
force left and the 4N upward force can be resolved
using Pythagoras. Answer = 5N
Practice/Application
4) A block of wood of mass 4kg rests on a slope,
inclined at an angle of 25° to the horizontal as
shown. Calculate:
The weight of the block
b) The normal reaction force of the plane acting on the
block.
c) The resultant accelerating force down the slope
d) The acceleration of the block down the slope
a)
Answers
4) There are several
a) Fg = mg => Fg = 40N
b) FN = Fg cosθ => FN = 36.6N
c) FR = Fg sinθ => FR = 16.9N
d) Fg = mg => a = 4.2m/s2
5) d
 Pg 73 # 1, 2, 4, 6, 7
Equilibrium
 Occurs when the net force on an object is zero.
 An object can move and still be at equilibrium.
 Neutral equilibrium(translational) – an object is at
equilibrium, it is then moved and then it is still at
equilibrium. A displacement results in another
equilibrium position. The net force acting on an object
is zero.
Newton’s Three Laws
 First Law - Inertia
 An object will remain at its state of rest or of constant
velocity unless acted upon by a force
Practice
1) For an object to be in translational equilibrium
a) It must be at rest
b) it must be moving with constant acceleration
c) No external force must be acting on it
d) The net force acting on it must be zero
2) A rain drop falling through air reaches a terminal
velocity before hitting the ground. At terminal
velocity, the frictional force on the raindrop is
a) Zero
b) Less than the weight of the raindrop
c) Greater than the weight of the rain drop
d) Equal to the weight of the raindrop
 A car is travelling along a level highway at a constant
velocity in a straight line. Air resistance is not
negligible. Draw the free-body diagram for this car.
Newton’s Second Law
 The rate of acceleration of an object is directly
proportional to the applied force and inversely
proportional to its mass.
 OR…. F=ma
Practice
4) A car mass 1100kg accelerates from 10m/s to 30m/s in
5.5s calculate the car’s engine force.
5) A boy on a bicycle, travelling at 12m/s applies the
brakes and comes to rest in a distance of 16m. If the
combined mass of the boy and bicycle is 70kg,
calculate the braking force.
6) An elevator of mass 800kg is supported by a thick
metal rope capable of withstanding a tension of
1200N. Calculate the tension in the rope when the
elevator is:
a) Not moving, between floors
b) Moving at constant velocity, of 2m/s upward
c) Moving at constant velocity, of 2m/s downwards
d) Accelerating upwards at a rate of 3m/s
e) Accelerating downwards at a rate of 3m/s
Newton’s Third Law
http://www.youtube.com/watch?v=8bTdMmNZm2M&f
eature=edu&list=PL772556F1EFC4D01C
 When two bodies A and B interact, the force that A
exerts on B is equal to the force that B exerts on A but
acts in the opposite direction.
7) A man pushes a car along a road. He exerts a force F
on the car. In this situation, what is the equal and
opposite force to F as referred to in Newton’s third
law?
8) Newton’s third law identifies pairs of forces that are
equal in magnitude. One of the forces acting on a
bird in flight is the gravitational force W downward.
What is the re-action force?