Movement and Motion 'Learning Objectives
Explain the distinction between scalar and vector quantities.
Distinguish between speed and velocity and define acceleration.
Calculate values using equations for velocity and acceleration.
Rate of Movement
The simplest thing we can measure is how fast an object is moving. You can calculate an
object's speed if you know the amount of time taken to move a certain distance:
“Speed: The rate of change of distance.”
Formula: speed(m/s) = distance(m)/time(s)
However, the calculation for speed will only tell you how fast an object is moving. Often
it is also vitally important to know in what direction this movement is taking the object.
When you include the direction in the information about the rate of movement of an
object, this is then known as velocity.
“Velocity: The rate of change of displacement, where the distance in a particular
direction is called the displacement.”
Formula: velocity(m/s) = displacement(m)/time(s)
Scalar and Vector Quantities
A scalar quantity is a measurement that has magnitude only, whereas a vector quantity
is a measurement that has both magnitude and direction.
Distance
Scalar
Vector
✅
❌
Scalar
Vector
Displacement
❌
✅
Speed
✅
❌
Velocity
❌
✅
Average and Instantaneous Speeds
In most journeys, it is unlikely that speed will remain constant throughout. As part of his
training programme, an athlete wants to keep a record of his speed for all races.
Average speed: Calculated by dividing the total distance for a journey by
the total time for the journey.
Instantaneous speed: The speed at any particular instant in time on a
journey, which can be found from the gradient of the tangent to a
distance-time graph at that time.
Skills Problem Solving
1. The athlete in fig B has taken 36 seconds from the start to reach the 300
m mark as shown. Calculate:
(a) his average speed during this 36 seconds
(b) his average velocity due north during this 36 seconds
(c) his average velocity due east during this 36 seconds.
Acceleration
Acceleration is defined as the rate of change of velocity. The equation defining
acceleration is:
“Acceleration: The rate of change of velocity.”
Formula: acceleration(m/s2) =
changeinvelocity(m/s)/timetakentochangethevelocity(s)
The vector nature of acceleration is very important. One of the consequences is that if
an object changes only the direction of its velocity, it is accelerating, while remaining at
a constant speed.
Motion Graphs �
One of the best ways to understand the movements of an object whilst on a journey is
to plot a graph of the position of the object over time. Such a graph is known as a
displacement-time graph.
Graph
Description
Displacement-time graph
Shows the position of an object over time.
Velocity-time graph
Shows the velocity of an object over time.
Acceleration-time graph
Shows the acceleration of an object over time.
Speed and Velocity from d-t Graphs
The gradient of the d-t graphs will tell us how fast the object was moving. Gradient is
found from the ratio of changes in the y-axis divided by the corresponding change on
the x-axis.
Formula: speed(m/s) = gradient = Δdistance(m)/Δspeed(s)## Distance and
Velocity from a Velocity-Time Graph
Key Concept: To find the distance travelled from a velocity-time graph, find the area
between the line and the x-axis.
Definition:
“Distance is the area between the line and the x-axis on a velocity-time graph.”
Area Under the Curve
Time (min)
Velocity (m/s)
Area
0-15
0.167
150 m
0-20
To find the distance travelled, calculate the area under the curve. In this case, the area is
a trapezium.
Gradient of a Velocity-Time Graph
Key Concept: To calculate the gradient of a velocity-time graph, divide a change in
velocity by the corresponding time difference.
Formula: gradient = Δv
Δt
This matches the equation for acceleration.
Acceleration from a Velocity-Time Graph
Key Concept: Acceleration is the rate of change in velocity.
Definition:
“Acceleration is the rate of change in velocity.”
Finding the Acceleration Due to Gravity
Method: Using multiflash photography or a video recording, measure the distance fallen
from the scale in each image. Create a carefully drawn distance-time graph to show a
curve as the object accelerates. Take regular measurements of the gradient by drawing
tangents to the curve.
Plotting Speeds on a Velocity-Time Graph
Key Concept: Plotting the speeds on a velocity-time graph should show a straight line,
as the acceleration due to gravity is a constant value.
Δv
Formula: g = Δt
Safety Note:
⚠ Persons with medical conditions such as epilepsy or migraine may be adversely
affected by multiflash photography.
Acceleration-Time Graphs �
Key Concept: Acceleration-time graphs show how the acceleration of an object
changes over time.
Definition:
“Acceleration-time graph shows how the acceleration of an object changes over
time.”
Example: Skydiver's Acceleration-Time Graph
Time (s)
Acceleration (m/s²)
0-10
9.81
10-20
...
...
Note: For a larger object falling for a long period, such as a skydiver, the acceleration will
change over time as air resistance increases with speed.
Subject Vocabulary �
Displacement-time graph: a graph showing the positions visited on a
journey, with displacement on the y-axis and time on the x-axis.
Velocity-time graph: a graph showing the velocities on a journey, with
velocity on the y-axis and time on the x-axis.
Gradient: the slope of a line or surface.
Let me know if this meets your requirements! �## Forces and Moments �
Resultant Force
The resultant force is the total force acting on a body when all the forces are added
together, accounting for their directions.
Free-Body Force Diagram
A free-body force diagram is a diagram showing an object isolated, and all the forces
that act on it are drawn in at the points where they act, using arrows to represent the
forces.
Principle of Moments �
Moment of a Force
The moment of a force is the tendency to cause rotation. It is calculated from:
“moment (Nm) = force (N) x perpendicular distance from the pivot to the line of
action of the force (m)”
Principle of Moments
The principle of moments tells us that if the total of all the moments trying to turn an
object clockwise is equal to the total moment of all moments trying to turn an object
anticlockwise, then it will be in rotational equilibrium.
Centre of Gravity ⚖
The centre of gravity is the point through which the weight of an object appears to act.
For a symmetrical object, it lies on every line of symmetry. The point of intersection of all
lines of symmetry will be the centre of gravity.
Finding the Centre of Gravity of an Irregular Rod
To find the centre of gravity of an irregular rod, we can use the principle of moments.
We can balance the rod on the edge of a thick metal ruler, and the centre of gravity
must lie above the ruler edge. Alternatively, we can use a set of hanging masses to
balance the rod more in the middle of the handle, and calculate the distance from the
pivot to the centre of gravity of the broom using the principle of moments.
Method
Description
Balancing on a
Balance the rod on the edge of a thick metal ruler, and the centre of gravity must lie
ruler
above the ruler edge.
Hanging
Use a set of hanging masses to balance the rod more in the middle of the handle,
masses
and calculate the distance from the pivot to the centre of gravity of the broom
using the principle!
Exam Hints
Note that the steps and layout of the solution in this worked example are
suitable for moments questions in the exam.
In equilibrium, principle of moments: sum of anticlockwise moments =
sum of clockwise moments.
Examples
1. What is the moment of a 252 N force acting on a solid object at a
perpendicular distance of 1.74 m from an axis of rotation of the object?
Answer: 252 × 1.74 = 438.48 Nm
2. A child and his father are playing on a seesaw, see fig H. They are exactly
balanced when the boy (mass 46 kg) sits at the end of the seesaw, 2.75 m
from the pivot. If his father weighs 824 N, how far is he from the pivot?
Answer: 824 × x = 46 × 9.8 × 2.75, solve for x.
3. The weight of the exercise book in the left-hand picture in fig B causes a
rotation so it moves towards the second position. Explain why it does not
continue rotating but comes to rest in the position of the second picture.
Answer: Because the moments are balanced, and the object is in equilibrium.
4. If the same set-up as shown in fig D was used again, but the toy car was
replaced with a banana weighing 1.4, find out where the banana would
have to be positioned for the beam to balance - calculate the new x3.
Answer: Use the principle of moments to calculate the new x3.## MOTION 1A.5
NEWTON'S LAWS OF MOTION '-
Newton's Second Law of Motion
Newton's Second Law of Motion tells us how much an object's motion will be changed
by a resultant force. For an object with constant mass, it is usually written
mathematically:
F = ma
Where:
F is the resultant force (in Newtons, N)
m is the mass of the object (in kilograms, kg)
a is the acceleration of the object (in meters per second squared, ms^-2)
This relationship allows us to calculate the acceleration due to gravity (g) if we measure
the force (F) accelerating a mass (m) downwards.
Experimental Verification of Newton's Second Law
The set-up shown in Fig. C measures the acceleration for various values of the resultant
force that acts on the trolley while keeping its mass constant (Table A). By plotting a
graph of acceleration against resultant force, a straight line will show that acceleration is
proportional to the resultant force.
Force (N)
Acceleration (ms^-2)
0.2
0.63
0.4
1.26
0.6
1.89
0.8
2.52
1.0
3.15
Graph of Results:
A straight line graph will show that acceleration is directly proportional to force.
Learning Tips
Straight-Line Graphs
Physicists always try to arrange their experimental data into graphs that produce a
straight best-fit line. This proves a linear relationship between the experimental
variables, and can also give us numerical information about the quantities involved.
The equation for a straight line is:
y = mx + c
Where:
m is the gradient of the straight line
c is the value on the y-axis where the line crosses it (the y-intercept)
If we plot experimental data on a graph and get a straight best-fit line, then this proves
the quantities we plotted on x and y have a linear relationship.
Newton's Third Law of Motion
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Newton's Third Law of Motion states that when an object A causes a force on another
object B, then object B causes an equal force in the opposite direction to act upon
object A.
Identifying Newton's Third Law
To find the two forces, remember they must always act on different objects and must
always have the same cause.
Object A
Object B
Force Direction
Skateboarder
Wall
Pushing away from wall
Wall
Skateboarder
Pushing towards wall
Football
Boot
Away from boot
Boot
Football
Towards boot
Analysis Checkpoint
1. In terms of Newton's laws of motion:
(a) Explain why this book will sit stationary on a table.
(b) Describe and explain what will happen if your hands then put an
upwards force on the book that is greater than its weight.
(c) Explain why you feel the book when your hands put that upwards
force on it.
2. (a) Calculate the gradient of the best-fit line on the graph in Fig. D and
thus work out the mass of the trolley that was accelerated in the first
investigation.
(b) State what quantity the gradient of the line on the graph in Fig. E
represents. Calculate the value of that quantity.
3. Calculate the acceleration in each of the following cases:
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(a) A mass of 12.0 kg experiences a resultant force of 785 N.
(b) A force of 22.2 N acts on a 3.1 kg mass.
(c) A 2.0 kg bunch of bananas is dropped. The bunch weighs 19.6 N.
(d) During a tackle, two footballers kick a stationary ball at the same
time, with forces acting in opposite directions. One kick has a force of
210 N, the other has a force of 287 N. The mass of the football is 430
g.
Subject Vocabulary
“Newton's First Law of Motion: an object will remain at rest, or in a state of uniform
motion, until acted upon by a resultant force.”
“Newton's Second Law of Motion: if an object's mass is constant, the resultant
force needed to cause an acceleration is given by the equation: F = ma”
“Newton's Third Law of Motion: for every action, there is an equal and opposite
reaction## Kinematics Equations '-”
Kinematics is the study of the description of the motion of objects.
Combining Equations A
We can combine the equations:
v = u + at and s = ut + (1/2)at^2
By substituting the first of these equations into the second, we get the combination
equation:
(u + v)/2 * t = s
We can use this equation to check again how high the stone was dropped from:
s = ut + (1/2)at^2 s = (0 x 3) + (1/2 x 9.81 x 3^2) = 44.1m
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Combining Equations B
We can combine the equations:
(u + v) = 2s/t and v = u + at
By substituting the first of these equations into the second, we get the combination
equation:
v = u + 2as
Check again what the stone's final velocity would be:
v = u + 2as = 0 + (2 x 9.81 x 44.1) = 29.4m/s
EXAM HINT: The kinematics equations are only valid if there is a constant acceleration. If
the acceleration is changing, they cannot be used.
Exam Questions
Question
Solution
1. What is the final velocity of a bike that starts at 4m/s and has zero
v = u + at = 4m/s + (0 x
acceleration acting on it for 10 seconds?
10s) = 4m/s
2. How far will the bike in question 1 travel in the 10-second time period?
s = ut + (1/2)at^2 = (4m/s x
10s) + (1/2 x 0 x 10s^2) =
40m
3. Calculate the acceleration in each of the following cases: ...
...
4. The bird in fig A drops the stone from a height of 88m above the
...
water surface. Initially, the stone has zero vertical velocity. How long will
it take the stone to reach the surface of the pond?
5. If the stone in fig A enters the water at 41.6m/s, and takes 0.65s to
...
travel the 3m to the bottom of the pond, what is its average
acceleration in the pond water?
Resolving Vectors �
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Resolving vectors is the process of splitting a vector into two components at right
angles to each other.
The Vertical Component of Velocity
“Resolution or resolving vectors: The reverse process of combining vectors. The
resolved pair of vectors will start at the same point as the original single vector.”
We can resolve the velocity vector into vertical and horizontal components:
vᵥ = v sin θ
vʰ = v cos θ
In the example of fig B, a basketball is thrown into the air at an angle of 40° to the
vertical. We can find the vertical and horizontal components of the velocity:
vᵥ = 4.2m/s sin 40° = 2.7m/s vʰ = 4.2m/s cos 40° = 3.2m/s
The Horizontal Component of Velocity
We can use the relationship:
vʰ = v cos θ
vʰ = 4.2m/s cos 40° = 3.2m/s
Resolving Vectors Calculations
Vector
Calculation
Solution
Velocity
vᵥ = v sin θ
vᵥ = 4.2m/s sin 40° = 2.7m/s
Velocity
vʰ = v cos θ
vʰ = 4.2m/s cos 40° = 3.2m/s
Alternative Resolution Angles
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The choice of direction of the right-angle pair, though, is arbitrary, and can be chosen to
suit a given situation.
Exam Questions
Question
Solution
1. (a) On graph paper, draw a velocity vector for a stone fired from a catapult at 45° to
...
the horizontal. ...
2. (b) Find the horizontal and vertical velocity components for this catapult stone by
...
calculation, and compare with your answers from part (a).
3. A javelin is thrown at 16m/s at an angle of 35° up from the horizontal. Calculate the
...
horizontal and vertical components of the javelin's motion.
4. A ladder is leant against a wall, at an angle of 28° to the wall. ...
...
Horizontal and Vertical Components of Force
When a force is applied at an angle, it can be resolved into horizontal and vertical
components. The horizontal component is parallel to the ground, while the vertical
component is perpendicular to the ground.
Example: Submarine Velocity
A submarine is moving at 240 m/s on a bearing of 125° from due north. To find its
velocity component due south and its velocity component due east, we can resolve the
velocity vector into horizontal and vertical components.
Component
Value
Velocity due south
vsin(125°)
Velocity due east
vcos(125°)
Resolving Vectors in Projectile Motion
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In projectile motion, the trajectory of an object is a parabola. The velocity of the object
can be resolved into horizontal and vertical components.
** Independence of Horizontal and Vertical Motion**
“"The actions in each of two perpendicular directions are wholly independent. This
means we can use Newton's laws of motion and the kinematics equations
separately for the horizontal and vertical motions of the same object."”
Kinematics Equations for Projectile Motion
Equation
Description
s = ut + (1/2)at^2
Vertical motion under gravity
v = u + at
Vertical velocity
v^2 = u^2 + 2as
Vertical velocity squared
Example: Stone Thrown Horizontally
A stone is thrown horizontally from a cliff with a velocity of 8.2 m/s. How much time is
the stone in flight? How far does it get away from the cliff by the time it lands?
Given
Value
Initial velocity
8.2 m/s
Acceleration due to gravity
9.81 m/s^2
Height fallen
60 m
Using the kinematics equations, we can calculate the time of flight and the horizontal
distance traveled.
Recombining Velocity Components
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To find the final velocity of the stone, we need to combine the horizontal and vertical
velocity components.
Component
Value
Horizontal velocity
8.2 m/s
Vertical velocity
-34.3 m/s
Final velocity
√(8.2^2 + 34.3^2) = 35.3 m/s
Horizontal Throws
In a horizontal throw, the object is thrown with an initial horizontal velocity. The
trajectory is a parabola, and the object will follow a symmetrical path.
Example: Boy Throws a Ball
A boy throws a ball vertically with an initial velocity of 4.8 m/s.
Question
Answer
How long is it before he catches it again?
1.54 seconds
What will be the ball's greatest height above the point of release?
23.3 m
Vertical Throws
In a vertical throw, the object is thrown with an initial vertical velocity. The trajectory is a
parabola, and the object will follow a symmetrical path.
Example: Basketball Throw
A basketball is thrown with a velocity of 6.0 m/s at an angle of 40° to the vertical.
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Question
Answer
If the hoop is 0.90 m above the point of release, will the ball
Yes
rise high enough to go in the hoop?
If the centre of the hoop is 3.00 m away, horizontally, from
Calculate the horizontal distance
the point of release, explain whether or not you believe this
traveled and compare to the
throw will score in the hoop.
distance to the hoop.
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