Topic 2 Mechanics
2.1 Kinematics
2.1.1 Define displacement, velocity, speed and acceleration.
Displacement- Defined as the change in position of an object. Displacement is a
quantity that has both direction and magnitude, it is a vector.
Measured in metres.
Speed- How far an object travels in a given time.
Rate of change of distance (ms-1)
Velocity- The vector version of speed. Tells us the magnitude of how fast an
object is moving and the direction in which it is moving. (ms-1)
Displacement
AverageVelocity =
TimeElapsed
Acceleration- Rate of change of velocity (ms-2)
ChangeInVelocity
AverageAcceleration =
TimeElapsed
2.1.2 Explain the difference between instantaneous and average
values of speed, velocity and acceleration.
Instantaneous values of velocity speed and acceleration are taken at a particular moment. In
any realistic situation they will be constantly varying and is not always meaningful or useful
to know this e.g. drivers need to keep their instantaneous speed below the speed limit.
Average velocity, speed and acceleration are taken over a certain period of time or distance.
E.g. Average speed between Porto and Lisbon = distance/time = 350/km/3 hours 30 min =
100 kmh-1
2.1.3 Outline the conditions under which the equations for uniformly
accelerated motion may be applied.
The following equations of motion may be applied to an object that has a constant
acceleration.
(v − u )
a=
t
s = ut + 12 at 2
v 2 = u 2 + 2as
v= final velocity/ ms-1)
u= initial velocity / ms-1
a= acceleration/ ms-2
s= displacement/m
t=time/s
2.1.4 Identify the acceleration of a body falling in a vacuum near the
Earth’s surface with the acceleration g of free fall.
,Free fall is the uniform acceleration (ignoring the effect of air resistance) in the vertical
direction of an object in a uniform gravitational field. All falling objects have the same
acceleration independent of their masses. On earth all objects accelerate towards the
ground at 9.81 ms-2.
Acceleration due to gravity can be measured in a number of
ways. In the arrangement that is shown a timer starts when
the ball is released form an electromagnet and stops when
the ball passes through a gate at the bottom. The
acceleration due to gravity can then be calculated using:
2s
g=
t2
Acceleration due to gravity can also be calculated using light
gates of ticker timers.
2.1.5 Solve problems involving the equations of uniformly
accelerated motion.
This question is about throwing a stone from a cliff.
Antonia stands at the edge of a vertical cliff and throws a stone vertically
upwards.
v = 8.0ms –1
Sea
–1
The stone leaves Antonia’s hand with a speed v = 8.0ms .
–2
The acceleration of free fall g is 10 m s and all distance measurements are
taken from the point where the stone leaves Antonia’s hand.
(a) Ignoring air resistance calculate
(i) the maximum height reached by the stone.
v 2 = u 2 + 2as
v2
s=
2s
, 82
s=
2 × 10
s = 3 .2 m
(2)
(ii) the time taken by the stone to reach its maximum height.
v−u
a=
t
v−u
t=
a
8
t=
10
t = 0 .8 s
(1)
The time between the stone leaving Antonia’s hand and hitting the sea is 3.0 s.
(b) Determine the height of the cliff.
Time to go form top of cliff to sea = 3.0 – (2 x 0.8) = 1.4s
s = ut + 12 at 2
s = (8.0 × 1.4) + ( 12 × 10 × 1.4 2 )
s = 21m
(3)
(Total 6 marks)
2.1.6 Describe the effects of air resistance on falling objects.
Air resistance acts upon all moving objects. As the velocity of an object increases the size of
the air resistance also increases.
Eventually the force of air resistance will equal the force of gravity. When this happens the
object will stop accelerating. It has reached its terminal velocity.
, The terminal velocity of an object is dependant on its shape. A feather has a lower terminal
velocity than a hammer.
2.1.7 Draw and analyse distance–time graphs, displacement–time
graphs, velocity–time graphs and acceleration–time graphs.
Displacement – time graphs
The object then
is stationary for
12 three seconds.
Object travels at a
10constant speed for the
first five seconds
dispalcement / m
8
The object returns to
6
its original location at a
faster speed.
4
2
0
0 2 4 6 8 10 12
time /s
The gradient of a displacement - time graph is the velocity.
2.1 Kinematics
2.1.1 Define displacement, velocity, speed and acceleration.
Displacement- Defined as the change in position of an object. Displacement is a
quantity that has both direction and magnitude, it is a vector.
Measured in metres.
Speed- How far an object travels in a given time.
Rate of change of distance (ms-1)
Velocity- The vector version of speed. Tells us the magnitude of how fast an
object is moving and the direction in which it is moving. (ms-1)
Displacement
AverageVelocity =
TimeElapsed
Acceleration- Rate of change of velocity (ms-2)
ChangeInVelocity
AverageAcceleration =
TimeElapsed
2.1.2 Explain the difference between instantaneous and average
values of speed, velocity and acceleration.
Instantaneous values of velocity speed and acceleration are taken at a particular moment. In
any realistic situation they will be constantly varying and is not always meaningful or useful
to know this e.g. drivers need to keep their instantaneous speed below the speed limit.
Average velocity, speed and acceleration are taken over a certain period of time or distance.
E.g. Average speed between Porto and Lisbon = distance/time = 350/km/3 hours 30 min =
100 kmh-1
2.1.3 Outline the conditions under which the equations for uniformly
accelerated motion may be applied.
The following equations of motion may be applied to an object that has a constant
acceleration.
(v − u )
a=
t
s = ut + 12 at 2
v 2 = u 2 + 2as
v= final velocity/ ms-1)
u= initial velocity / ms-1
a= acceleration/ ms-2
s= displacement/m
t=time/s
2.1.4 Identify the acceleration of a body falling in a vacuum near the
Earth’s surface with the acceleration g of free fall.
,Free fall is the uniform acceleration (ignoring the effect of air resistance) in the vertical
direction of an object in a uniform gravitational field. All falling objects have the same
acceleration independent of their masses. On earth all objects accelerate towards the
ground at 9.81 ms-2.
Acceleration due to gravity can be measured in a number of
ways. In the arrangement that is shown a timer starts when
the ball is released form an electromagnet and stops when
the ball passes through a gate at the bottom. The
acceleration due to gravity can then be calculated using:
2s
g=
t2
Acceleration due to gravity can also be calculated using light
gates of ticker timers.
2.1.5 Solve problems involving the equations of uniformly
accelerated motion.
This question is about throwing a stone from a cliff.
Antonia stands at the edge of a vertical cliff and throws a stone vertically
upwards.
v = 8.0ms –1
Sea
–1
The stone leaves Antonia’s hand with a speed v = 8.0ms .
–2
The acceleration of free fall g is 10 m s and all distance measurements are
taken from the point where the stone leaves Antonia’s hand.
(a) Ignoring air resistance calculate
(i) the maximum height reached by the stone.
v 2 = u 2 + 2as
v2
s=
2s
, 82
s=
2 × 10
s = 3 .2 m
(2)
(ii) the time taken by the stone to reach its maximum height.
v−u
a=
t
v−u
t=
a
8
t=
10
t = 0 .8 s
(1)
The time between the stone leaving Antonia’s hand and hitting the sea is 3.0 s.
(b) Determine the height of the cliff.
Time to go form top of cliff to sea = 3.0 – (2 x 0.8) = 1.4s
s = ut + 12 at 2
s = (8.0 × 1.4) + ( 12 × 10 × 1.4 2 )
s = 21m
(3)
(Total 6 marks)
2.1.6 Describe the effects of air resistance on falling objects.
Air resistance acts upon all moving objects. As the velocity of an object increases the size of
the air resistance also increases.
Eventually the force of air resistance will equal the force of gravity. When this happens the
object will stop accelerating. It has reached its terminal velocity.
, The terminal velocity of an object is dependant on its shape. A feather has a lower terminal
velocity than a hammer.
2.1.7 Draw and analyse distance–time graphs, displacement–time
graphs, velocity–time graphs and acceleration–time graphs.
Displacement – time graphs
The object then
is stationary for
12 three seconds.
Object travels at a
10constant speed for the
first five seconds
dispalcement / m
8
The object returns to
6
its original location at a
faster speed.
4
2
0
0 2 4 6 8 10 12
time /s
The gradient of a displacement - time graph is the velocity.
