2. Acceleration, deceleration, velocitytime graph interpretation, calculations and
problem solving
Doc Brown's Physics Revision
Notes
Suitable for GCSE/IGCSE Physics/Science courses or
their equivalent
This page will help you answer questions e.g. How do you define acceleration?
How do you interpret velocitytime
graphs? How to do acceleration formula
questions? What to we mean by terminal velocity?
Subindex for this page
(a)
Brief reminders from
section 1. on distancetime graphs
(b)
Introducing
velocitytime graphs and acceleration
(c)
Example questions
based on graph interpretation
(not mathematical calculations)
(d)
Acceleration formula,
calculations, interpreting velocitygraphs (a
= Δv ÷ Δt)
(e)
More advanced
calculations involving uniform acceleration (using v^{2}  u^{2}
= 2as)
(f)
Forces and
circular motion  including acceleration and centripetal force
(a) Brief reminders from section 1. on distancetime
graphs
Distance  time graphs
Graph
shows
acceleration, speeding up
Graph
shows
zero speed/velocity
Graph
deceleration,
slowing down
Graph
shows
uniform or constant speed/velocity, not acceleration
DO NOT confuse these graphs with the
velocitytime graphs below!
TOP OF PAGE
and subindex
(b) Introducing velocitytime
graphs and acceleration
(DO NOT confuse these graphs with the
ones above!)
Acceleration is about change in speed or
velocity of an object and is not the same as speed or velocity.
The interpretation of the seven graphs above,
by considering the gradient  positive, negative or zero,
and their mathematical sign if the values
of acceleration (+) or deceleration () are used in calculations.
acceleration is the rate of change of
velocity with time
acceleration = change in velocity ÷ time
taken to effect the velocity change
a
(m/s^{2})
= Δv (m/s)
÷ Δt (s) (actual numerical calculations are
dealt with further down the page)
A comparison of speed/velocity  time graphs
Graph
curves upwards, velocity is increasing (+ve), but also shows
increasing acceleration, speeding up at an increasing rate.
A comparison of speed/velocity  time graphs
Graph
is
flat, shows
constant velocity/speed (+ve), zero acceleration. Cruise control operating in a
car.
A comparison of speed/velocity  time graphs
Graph
shows
decreasing acceleration (+ve), BUT not slowing down!, still speeding up but at a
decreasing rate.
This is observed as you pull away in a car up
to a maximum speed dictated by the speed limit.
A comparison of speed/velocity  time graphs
Graph
is
linear, shows
constant or uniform acceleration (+ve), constant rate of increasing velocity.
A comparison of speed/velocity  time graphs
Graph
is
linear, shows
constant or uniform deceleration (ve), velocity decreasing, constant rate of deceleration.
A comparison of speed/velocity  time graphs
Graph
shows
decreasing deceleration (ve), velocity decreasing, but decelerating at an increasingly slower rate.
A comparison of speed/velocity  time graphs
Graph
shows
increasing deceleration (ve), velocity decreasing, but slowing down at an increasingly faster rate!
You can demonstrate this by gradually
increasing the pressure on the brake pedal of a car and bringing it to sudden
halt.
The seven graphs above illustrate of what you
might see at any point on a velocitytime graph, BUT, in reality, any
speed/velocitytime graph will be far more complicated than any one of these
simple graphs. So, you may find a graph with all seven types of gradient in just
one speed/velocitytime graph, like the velocitytime graph below which could
represent a car journey.
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and subindex
(c)
Example questions based on graph interpretation
(not involving mathematical calculations)
Q1.1 The
graph below describes the velocity profile of a car journey.
Analyse the graph sections indicated below and describe the
motion of the car at the various stages of the journey:
ab: increasing acceleration,
bc: linear/uniform/constant acceleration,
cd: decreasing acceleration,
de: constant velocity, zero/no acceleration or
deceleration
ef: increasing deceleration
fg: linear/uniform/constant deceleration,
gh: decreasing deceleration
Q1.2 The sketch above illustrates the design of a roller coaster ride.
Using the numbers (1) to (9) and comment
on energy store situations or changes ...
(a) Give two places where acceleration is
taking place.
(4) and (5) accelerating
downhill, conversion of
gravitational potential energy (GPE) to kinetic energy (KE)
(b) At what points does the car have its
maximum GPE?
at (3) and (7), the highest
points you have the maximum in GPE
(c) At which point are you getting the
greatest increase in GPE?
(1) => (2) => (3), the largest
climb of the circuit.
(d) At which points do you get
deceleration?
Climbing up at (6) as KE store
decreases and GPE store increases.
(e) At which point do you get the maximum
change of KE to increase a thermal energy store?
The braking zone <==(9)
Q1.3
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and subindex
(d) Acceleration formula, problem solving calculations and
interpreting velocitygraphs
Acceleration is about change in speed or
velocity of an object and is not the same as speed or velocity.
Acceleration has its own defined formula,
which expresses a change in velocity in a defined time interval.
acceleration = change in velocity ÷ time
taken to effect the velocity change
a
(m/s^{2}) = Δv
(m/s) ÷ Δt (s)
a = acceleration in metres per second squared m/s^{2},
Δv = change in speed/velocity in
metres/second m/s
Δt = the time taken for the
speed/velocity change in seconds, s
The formula for acceleration can also be
written as:
a = (v  u) ÷ t
= Δv ÷ t
where v = final speed/velocity, u = initial speed/velocity, t =
time taken to change from velocity u to v
Rearrangements of acceleration formula:
If a = Δv ÷ t
then Δv
= a x
t and
t = Δv ÷ a
Velocity  time graphs  illustrating the motion of objects A to F
From the graphs you can deduce several
things:
From the gradient you can obtain the
acceleration or deceleration (negative acceleration).
From the area under the 'triangles'
or 'rectangles' of the velocity  time graph you can deduce the distance
travelled  the displacement equals the area under a velocity time
graph.
In terms of units, the
distance calculation is equivalent to time x (distance / time)
e.g. = time x speed = s x m/s =
m
Graph (1) Linear or uniform
acceleration
Both graph lines represent a linear
increase in velocity AND the acceleration is constant.
From the equation a
(m/s^{2}) = Δv
(m/s) ÷ Δt (s)
For object A: acceleration = (50
 0) ÷ (25  0) =
2.0 m/s^{2}
For object B: acceleration = (50
 30) ÷ (20  0) =
1.0 m/s^{2}
Note that the steeper the
gradient, the greater the acceleration, the greater the rate of
'speeding up'.
Examples of distance travelled
calculations from graph (1)
Here you are only dealing with an
area of a triangle.
For object A between 0 and
20 seconds the distance travelled = (20 x 40) / 2 =
400 m
Because you are working with
the area of a triangle, don't forget to halve the calculation!
For object B between 0 and
20 seconds the distance travelled = {20 x (50  30)} / 2 =
200 m
Graph (2) Constant velocity  zero
acceleration
The graph lines are horizontal,
indicating there is no change in velocity, therefore zero acceleration.
Graph line C for object C
represents a constant velocity of 20 m/s.
Graph line D represents an object
D moving at twice the speed of C at 40 m/s.
Examples of distance travelled
calculations from graph (2)
You are only dealing with a
rectangle.
e.g. distance travelled by
object C between 5 and 15 seconds = (15  5) x (200) =
200 m
and distance travelled by
object D between 0 and 25 seconds = (25  0) x (40  0) =
1000 m
Graph (3) Deceleration  negative
acceleration
Both graph lines represent a linear
decrease in velocity AND the deceleration (negative acceleration) is
constant.
From the equation a
(m/s^{2}) = Δv
(m/s) ÷ Δt (s)
For object E: acceleration = (0 
40) ÷ (25  0) = 1.6
m/s^{2}
For object F: acceleration = (30
 50) ÷ (25  0) =
0.8 m/s^{2}
Remember, if you express the
answers as 1.6 m/s^{2} and 0.8 m/s^{2}, it might be
a good idea to say 'deceleration' or 'slowing down' too!
Note that the steeper the
gradient, the greater the deceleration, the greater the rate of
'slowing down'.
Examples of distance travelled
calculations from graph (3)
Again, like graph (1), it only
involves the area of a triangle.
Distance travelled by object E
between 0 and 25 seconds = 25 x (40  0) / 2 =
500 m
Distance travelled by object F
between 10 and 25 seconds = (25  10) x (42  30) / 2 =
90
m
Note on use of words:
You can think of deceleration is negative
acceleration, so when a moving object is slowing down the rate of change in
speed/velocity with time
is negative.
Therefore you can think of acceleration as the positive rate of change of speed/velocity.
But, you may find that the deceleration
isn't used, and a negative sign used with the word acceleration.
So watch out for mathematical signs and
how the words acceleration and deceleration are used.
Example calculations based on the acceleration formula
and velocitytime graph interpretation
Q2.1 A car
accelerates from 10 m/s to 30 m/s in 15 seconds.
Calculate the acceleration of the car.
a
(m/s^{2}) = Δv
(m/s) ÷ Δt (s) ÷ a
If u and v are the initial and final
velocities
a = (v  u) / t = (30  10) / 15 = 20 / 15 =
1.3 m/s^{2}
(2 s.f.)
Q2.2 A
train accelerates at 0.50 m/s^{2} for 30 seconds.
What is the increase in speed of the train?
a = Δv ÷ t so
Δv = a
x
t
increase in speed = 0.5 x 30 =
15
m/s (2 s.f.)
Q2.3 If a
car accelerates at 0.30 m/s^{2} from a standing start, how long will it
take to attain a speed of 21 m/s?
a = Δv ÷ Δt so
Δt = Δv ÷ a = (v  u) ÷ a,
u and v are the initial and final velocities
time taken = (21  0) ÷ 0.30 =
21 s (2
s.f.)
Q2.4 A car brakes sharply from moving at 30 m/s to 10 m/s in 4.0
seconds.
Calculate the deceleration of he car.
a = Δv ÷ Δt
If u and v are the initial and final
velocities
a = (10  30) / 4.0 = 20 / 4 =
5.0 m/s^{2}
(2 s.f.)
Note the minus sign for the acceleration
 because the car is slowing down!
You can then say the deceleration is
5.0 m/s^{2}
(2 s.f.)
Q2.5 The
graph below summarises the first 100 seconds of a cyclist's race.
When calculating displacement (distance
travelled), with this velocity  time graph, both triangles and rectangles are
involved.
(a) (i) What is the initial uniform acceleration of the cyclist in
this time period?
The initial acceleration is from 0 to 40 seconds.
speed = total change in speed / time taken = (10  0) / (40
 0) = 10 / 40 = 0.25 m/s^{2}
(2 s.f.)
(ii) How far has the cyclist travelled in this period?
From a speed/velocitytime graph you
calculate distance covered from the area under the graph for the time period
involved.
Area under the graph from 0 to 40 seconds = 10 x 40 / 2 =
200 m (its easy if the graph is linear)
(b) At which point is the cyclist's acceleration the greatest?
and calculate the acceleration at this point.
The steeper the gradients the greater the acceleration.
The gradient is steepest from
90 to
100 seconds, so that is the time period of greatest acceleration.
a = ∆v / ∆t = (25  20) / (100  90) = 5
/ 10 =
0.5 m/s^{2}
(c)
Q2.6 The
graph below summarises a 100 minute commuter train journey, but with some signal
delays!
Watch out for min ==> hour conversions!
(a) At what time or times is the train
moving at constant speed? Explain your reasoning.
From the 20th to 40th minute
AND from the 50th to 70th minute. The graph is horizontal,
indicating constant speed (of 100 km/hour and 200 km/hour).
(b) Where is the greatest acceleration
and what is its value in km/hour^{2}?
Steepest gradient is from the 40th
to the 50th minute.
Change in velocity = 200  100 = 100
km/hour
Time involved = 50  40 = 10 mins,
10/60 = 0.167 hours
gradient = 100 / 0.167 = 100 / 0.167
= ~600
km/hour^{2}
(c) How far does the train travel in the
first 20 minutes?
time = 20 / 60 = 0.333 hours, speed
change = 100  0 = 100 km/hour.
area under graph = 100 x 0.333 / 2 =
16.7 km, ~17 km
Q2.7
A car travelling at 36 m/s skids off the road and hits a wall.
(a) If the car is brought to a halt in 2.0 seconds, what is
the average deceleration of the car?
final velocity v = 0, initial velocity u = 36 m/s
acceleration
a = (v  u) ÷ t = Δv ÷ t
= (0  36) / 2.0 = 18 m/s^{2}
So the average deceleration is 18 m/s^{2}
Note: The acceleration due to gravity
is 9.8 m/s^{2}, so in this crash you body would experience a
force of nearly '2G', not good for your body!
(b)
Q2.8 The graph below profiles the speed of a cyclist at the start of
a race.
NOT a
linear graph!
You get the distance travelled by calculating or measuring
off the graph, the area under the graph for the specified time interval. To
help you appreciate the method, I've highlighted two areas in yellow to be
estimated to answer the two questions below.
Step 1 is to work out the distance relating to one small
square.
From the 'bold' graph lines larger square you have 5 x 5 =
25 small squares = 2 m/s x 20 s = 40 m,
so each small square = 40 / 25 = 0.625 m, so
from this you can estimate the areas and distances.
(a) How far does the cyclist travel in the first 20 seconds
of the race? (give your answers to 2 s.f.)
010s ~ 2 squares, 1020s ~9, total ~11 squares = ~11 x
0.625 = 6.88, distance travelled
~6.9
m
(b) How far does the cyclist travel between the 30th and
40th second of the race.
total small squares = 25 + 22
(25~3) + 5.5 = ~52.5, distance travelled = 52.5 x 0.625 = ~32.8 =
~32
m
Do
you agree with my estimations?
Q2.9
A van is travelling at 20 m/s has its brakes applied for 5.0 seconds.
If this produces a deceleration of
1.5 m/s^{2}, calculate the velocity of the van after braking.
a = Δv ÷ Δt
Δv
= (v  u) = a x
t = 1.5 x 5.0 = 7.5
v  u = 7.5, v  20 =
7.5, v = 7.5 + 20 =
12.5 m/s
Q2.10 If an object is fired vertically up in the air from ground
level,
(a) Describe the energy store
changes taking place and any assumptions you are making in
your arguments. Assume the object leaves from, and returns to, the
ground level.
When the object is thrown upwards
it is given an initial maximum kinetic energy store.
As it rises the velocity
decreases, its kinetic energy store (KE) decreases and its
gravitational potential energy store (GPE) increases proportionately
 this assume no other energy transfers are occurring.
This ignores the friction effect
of air resistance  some KE is lost as thermal energy to increase
the thermal energy store of the surroundings.
The decrease in velocity is due
to the decelerating effect of the Earth's gravitational field.
When it reaches its maximum
height, the object has its maximum GPE.
The object then accelerates back
down to the ground, again due to the Earth's gravitational field.
At the same time, as the velocity
increases, the object's GPE store is converted to its KE store and
it hits the ground at its maximum velocity and with its maximum KE.
This assumes no air resistance.
(b) Sketch the graph of velocity
versus time for the whole flight.
There are two graph formats you
can draw to illustrate the change in velocity with time and both are
illustrated below.
Graph A is constructed in two halves.
From the maximum velocity x, the left half represents the
decrease in velocity as the object's motion is slowed down by the
Earth's gravitational field.
You can think of this as a
negative acceleration i.e. deceleration  a negative gradient (a 
m/s^{2}).
The right half represents the
increase in velocity as the object falls under gravity, attaining
the maximum velocity x on hitting the ground.
You can think of this as a
positive acceleration  a positive gradient (a + m/s^{2}).
Graph A ignores the mathematical
convention of a minus velocity on change of direction.
The bottom of the V, at time t, represents
zero velocity as the object's motion is on the point of reversing
direction to fall down.
Graph B is constructed as one
continuous line.
This uses both positive and
negative values of velocity, to represent the change in direction.
Therefore the velocity changes
from +x to x in a time of 2 x t, t being the point of zero velocity
(c) Can you suggest a value for the
gradients of the graph?
The gradients in each case would have
the value of the Earth's gravitational acceleration/deceleration
constant of 9.8 m/s^{2}.
Q2.11
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and subindex
(e)
More advanced calculations involving uniform acceleration using
v^{2}  u^{2}
= 2as
(this section might not be required for your course)
Constant acceleration is known as linear uniform
acceleration.
A good example is an object free falling
in a gravitational field e.g. it is the acceleration due to gravity is 9.8 m/s^{2}
at the Earth's surface.
You can use the following formula to do
calculations based on a uniform acceleration situation.
An equation for uniform acceleration is
...
(final speed (m/s))^{2} 
(initial speed (m/s))^{2}
= 2 x acceleration (m/s^{2}) x distance (m)
v^{2}  u^{2}
= 2as, v = final velocity,
u = initial velocity,
a = acceleration,
s
= distance travelled or displaced
Rearrangements to calculate the various
object variables in the above equation:
(units above, and remember in algebraic equations,
if you 'change the side you
change the sign, or you change all the signs of the expression)
(i) v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
(ii) v^{2}  u^{2} = 2as,
u^{2} = 2as  v^{2}, u^{2} = 2as
+ v^{2},
u =
√(v^{2}  2as)
(iii) v^{2}  u^{2} =
2as, 2as = v^{2}  u^{2},
a = (v^{2}  u^{2})
÷ 2s
(iv) v^{2}  u^{2} = 2as,
2as = v^{2}  u^{2},
s
= (v^{2}  u^{2}) ÷ 2a
Examples questions based on uniform
acceleration
Q3.1 An
initially stationary car accelerates uniformly at a rate of 1.50 m/s^{2}.
(a) What is the speed of the car after a distance travelled
of 100 m?
v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
Since u = 0 (initially car is
stationary) and s = 100 m
v = √(2 x 1.5 x 100) = √300 =
17.3 m/s
(3 s.f.)
(b) If at 100 m the driver presses harder on the accelerator
pedal to increase the acceleration to 2.00 m/s, what speed will the car be
doing after travelling a total of 300 m from the start?
v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
u = 17.3 m/s (from part (a)), s = 300  100 (you must
subtract 100 m because that's where new acceleration started
v = √{(2 x 2.0 x 200) +
(17.3^{2})} = √(800 + 300) = √1100 =
33.1
m/s (3 s.f.)
Note the use of ( ) AND { }
brackets, you really must take care about what follows the overall
square root sign √.
Q3.2
A jet rocket plane is launched from a converted jet airliner with an
acceleration of 10.0 m/s^{2}.
From radar tracking it is found that the rocket plane
achieved a speed of 300 m/s after travelling for 1875 m.
At what speed was the launch aircraft travelling?
v^{2}  u^{2} = 2as,
u^{2} = 2as  v^{2}, u^{2} = 2as
+ v^{2},
u =
√(v^{2}  2as)
Initially both the launch aircraft and
the rocket plane are travelling at the same speed u.
v = 300 m/s, a = 10 m/s^{2}, s =
1875 m
u = √(v^{2}  2as) =
√{(300^{2})  (2 x 10.0 x 1875)}
u = √(90000  37500) = √52500 =
229 m/s
(3 s.f.)
Q3.3 A car
is moving at a speed of 20 m/s when the driver presses on the accelerator pedal
and after a distance of 400 m the car is travelling at 30 m/s. Assuming it is
uniform, calculate the acceleration of the car.
v^{2}  u^{2} =
2as, 2as = v^{2}  u^{2},
a = (v^{2}  u^{2})
÷ 2s
s = 400 m, u = 20 m/s, v = 30 m/s (u and
v being the initial and final velocities)
a = (30^{2}  20^{2}) /
(2 x 400) = (900  400) / 800 = 0.625 =
0.63 m/s^{2}
(2 s.f.)
Q3.4
A rocket is moving vertically upwards and leaves the launch silo with a velocity
of 10.0 m/s. Booster rockets are immediately fired to produce an acceleration of
4.00 m/s^{2}.
At what height (s) above the launch pad, will the rocket
attain a velocity of 200 m/s (give your final answer in km).
(iv) v^{2}  u^{2} = 2as,
2as = v^{2}  u^{2},
s
= (v^{2}  u^{2}) ÷ 2a
a 4.00 m/s^{2}, u = 10.0 m/s, v = 200 m/s
(u and v being the initial and final velocities)
s = (200^{2}  10^{2}) /
(2 x 4) = (40000  100) / 8 = 39900 / 8 = 4987.5 m =
4990
m or
4.99 km (3 s.f.)
Note: To fire an object
vertically from a 'powerful gun', to completely escape the Earth's
gravitational field, requires an 'escape velocity' of 11 km/s at the end of
the 'gun barrel'! However, rockets don't function like a gun. They have
continuous thrust force from the rocket engines which will sustain a steady
upward movement against the force of gravity.
Q3.5 An
object is dropped from a height of 10 m above the ground. Ignoring air
resistance, if the acceleration due to gravity is 9.8 m/s^{2}, at what
theoretical velocity does the object hit the ground?
v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
u = 0, v = ?, a = 9.8 m/s2, s = 10 m
since u = 0, v = √(2 x 9.8 x 10) = √196 =
14 m/s (2 s.f.)
Note: (i) Not a good idea to fall
when rock climbing, falling from ~31 ft and hitting the ground at 14 m/s.
This will do a lot of damage, but no problem if all safety devices in place!
BUT, imagine a 100 m sprinter running into a brick wall at an average speed
of ~10 m/s!
(ii) The general formula for calculating
this 'fall' velocity, since u is zero, is:
v = √(2as + u^{2}) =
√(2as) = √(2 x 9.8 x height) = √(19.6 x height above ground)
e.g. if you fall off a ladder at
height of just 2 m above the ground (~6 ft) you hit the ground at:
v = √(2as) = √(2 x 9.8 x 2) =
√39.2 = 6.3 m/s.
Comparing this with the speed of
a sprinter, its hardly surprising the serious injuries you can
suffer from a relatively short fall.
Q3.6
A an object is dropped from a certain height above the ground.
If the object hits the ground at 10 m/s,
from what height was it dropped?
v^{2}  u^{2} = 2as,
rearranging s
= (v^{2}  u^{2}) ÷ 2a
s = height, v = 10 m/s, u = 0 m/s, a
= 9.8 m/s^{2} (gravitational acceleration near Earth's surface)
height = s = (10^{2}
 0^{2}) ÷ (2 x 9.8) = 100 ÷ 19.6 =
5.1 m
Q3.7
An object is fired vertically up into the air with an initial velocity of 20.0
m/s.
Ignoring air resistance, and taking the
acceleration as 9.80/ms^{2}, calculate the theoretical maximum
height the object attains.
v^{2}  u^{2} = 2as,
rearranging s
= (v^{2}  u^{2}) ÷ 2a
s = (0^{2}  20^{2})
÷ (2 x 9.8)
s = 400 ÷ 19.6 =
20.4 m (to 3 sf)
Q3.8 A falling object is travelling at 10 m/s
when it is 10 m above the ground.
If gravitational acceleration is 9.8
m/s^{2}, at what velocity will it hit the ground?
v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
Substituting in the rearranged
equation:
v = √{(2 x 9.8 x 10) + 10^{2})
v = √(196 + 100) = √296
final velocity on hitting the
ground =
17 m/s (2 sf)
Q3.9
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and subindex
(f) Forces and
circular motion  including acceleration and centripetal force
This section was adapted, reedited and extended from the web page including
a section on Gravity and circular
motion.
Velocity is a vector quantity, it has both size (the
speed) and direction.
If either the speed or direction changes, you have a
change in velocity  you have an acceleration!
With this in mind, imagine whirling a
conker around on the end of a piece of string, the moon orbiting Earth
or planets orbiting the Sun.
What velocity are we dealing with? What force are we
dealing with?
When an object goes round in a circle
at a constant speed its direction is continually changing.
Even though the speed is constant,
because direction changes, the
velocity
(a vector)
is continually changing.
Therefore, if the velocity is
changing, you must be dealing with an
acceleration, even though the speed is
constant.
To maintain this accelerating
circular motion, there must be a force operating to maintain the
circular path.
This is called the
centripetal force  it can be the tension in a string
as you whirl an object around or gravity holding some large object in
orbit.
The direction of acceleration is
inwards, the same direction as the centripetal force, and at 90^{o}
to the direction of motion.
(You don't have to know this, but the
acceleration of an object moving at a constant speed in a circle
is v^{2}/r)
circular motion  velocity & centripetal force
To keep a body moving in a circle there must be a force
directing it towards the centre.
This is called the centripetal force and produces the continuous change in direction
of circular motion.
Even though the speed may be constant, the object is
constantly
accelerating because the direction is constantly changing via
the circular path  i.e.
the velocity is constantly changing (purple arrows, on the diagram).
For an object to be accelerated, it
must be subjected to a force that can act on it  Newton's 1st law of
motion.
Here the resultant centripetal force
is acting towards the centre, so always directing the object to 'fall'
towards the centre of motion (blue arrows on the diagram).
But the object is already moving, so
the force causes it to change direction.
SO, the actual circular path of motion is determined by
the resultant centripetal force (black arrows and circle) and the
circling object keeps accelerating towards what it is orbiting.
The centripetal force stops the
object from going off at a tangent in a straight line.
Swinging something round on a string.
When you swing something round on the end of a string, the tension in the string is the centripetal force.
Imagine whirling a conker round on
the end of a string.
You yourself feel this force of tension as the 'pull' in
the string (I've marked in a black line to represent the string).
If you could use a fast action camera to monitor the
motion and the string broke, you would observe the object would fly off
at the precise tangent to the circular path and in a straight line of
constant velocity  the result resultant of Newton's 1st law!
Since gravity and air friction act on the object, you do
have to keep on 'inputting' kinetic energy to keep it swinging round.
The centripetal force will vary with the mass of the object,
the speed of the object and the radius of the path the object takes.
The same arguments on circular motion apply to the movements
of planets around a sun, a moon around a planet and a satellite orbiting a
planet. The orbits are usually elliptical, rarely a perfect circle, but the
physics is the same.
In these cases, it is the force of gravitational attraction
that provides the centripetal force and it acts at right angles to the
direction of motion.
You should also realise that they are moving through empty
space (vacuum), so there are no forces of friction to slow the object down.
This is why the planets keep going around
the Sun and the moon keeps going around the Earth.
When satellites are put into orbit they
are given just the right amount of horizontal velocity so that the
resultant centripetal force of gravity keeps the satellite in its a
circular orbit.
You can vary this horizontal velocity to
position satellites at different distances above the Earth's surface.
TOP OF PAGE
and subindex
Motion and associated forces notes index (including Newton's Laws of
Motion)
1. Speed and velocity  the relationship between
distance and time, distancetime graphs gcse physics
2. Acceleration, velocitytime graph interpretation and calculations,
problem solving
gcse physics revision notes
3. Acceleration,
friction, drag effects and terminal velocity experiments
gcse physics revision notes
4. Newton's First, Second and Third Laws of
Motion, inertia and F = ma calculations
gcse physics revision notes
5. Reaction times stopping distances and example
calculations
gcse physics revision notes
6. Elastic and nonelastic collisions, momentum
calculations and Newton's 2nd law of
motion
gcse physics notes
IGCSE revision
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