Tuesday, 29 November
5.1.2 Kinematics part 1 ‘Graphs’ 2011
You should be able to:
(a) define displacement, speed, velocity and acceleration.
(b) use graphical methods to represent distance travelled,
displacement, speed, velocity and acceleration.
(c) find the distance travelled by calculating the area under a
speed-time graph.
(d) use the slope of a displacement-time graph to find
velocity, and of a distance-time graph to find speed.
(e) use the slope of a velocity-time graph to find
acceleration.
(h) interpret displacement-time and speed-time graphs for
motion with non-uniform acceleration.
Write out definitions for the following terms:
distance,
displacement,
speed,
velocity and
acceleration
These are Definitions for linear motion (i.e. in a
straight line)
Definitions for linear motion (i.e. in a straight line)
distance is how far you travel between any two points by
any route. It is a scalar quantity.
displacement is the minimum “as the crow flies” distance
between two points. It is a vector quantity, so it has
direction.
speed is how fast you go, the rate of change of distance.
velocity is rate of change of displacement. It must have
a direction.
Acceleration can be used as both a vector and a scalar
quantity. It is the rate of change of speed or velocity.
Distance-time graphs. Match the graphs to the descriptions.
moving at a slow speed
staying still at the start
time (s)
time (s)
moving and increasing speed
staying still at a short distance from the start
moving at a fast speed
time (s) time (s)
time (s)
Look at the animation and draw two lines on a
distance-time graph to show the motion
Press the space bar to see the graph.
Which of the statements below is definitely true about
the motion of the cars below.
The red car moves
at a steady speed. The blue car has
The green car the highest speed.
accelerates
The blue car the least.
accelerates The red car
the least. slows down.
The red car is
The red car faster than the The green car moves
accelerates. green car. at a steady speed.
The red car moves The red car The red car The red car is
at a steady speed. slows down. accelerates. faster than the
green car.
The red car neither accelerates nor slows down but
moves at a steady speed.
Initially it goes faster than the green car but the green
car catches up with it meaning that the green car is going
faster at the end.
The blue car
accelerates The blue car has
the least. the highest speed.
The blue car has the highest velocity at
the end but it also accelerates the most.
The green car
The green car moves
accelerates
at a steady speed.
the least.
The green car speeds up but doesn‟t
accelerate as much as the blue car.
However you could argue that the red car
with zero acceleration has a lower
acceleration !
B
C
A
What cars are described by the graphs A, B and C ?
Displacement, Velocity and Acceleration graphs
There are three sorts of graphs:
Displacement-time
Velocity-time
Acceleration-time.
Information from a displacement-time graph.
decreasing
steady
velocity
velocity
Displacement (m)
Copy the graph and
time (s) sketch a velocity-
time graph for the
motion in each of
the different
increasing steady position shown
velocity velocity beside each label .
The motion of the blue ball is tracked with a ticker tape
timer. Write a description of the motion, copy the diagram
and complete the displacement, velocity and acceleration
graphs below. Leave a space for a title.
Example 1/6
Again, write a description of the motion shown above, copy
the diagram and complete the displacement, velocity and
acceleration graphs below.
Example 2/6
Describe the motion and complete the graphs
Example 3/6
Describe the motion and complete the graphs
Example 4/6
Describe the motion and complete the graphs
Example 5/6
Describe the motion and complete the graphs
Example 6/6
Use these titles to label each diagram
Rightward motion with rightward acceleration
Leftward motion with leftward acceleration
Constant velocity leftwards
Rightward motion with leftward acceleration
Constant velocity rightwards
Leftward motion with rightward acceleration
5.1.2 Kinematics part 2 ‘suvat equations’
You should be able to:
(f) derive, from the definitions of velocity and of
acceleration, equations which represent uniformly
accelerated motion in a straight line.
(g) use equations which represent uniformly accelerated
motion in a straight line, including falling in a uniform
gravitational field without air resistance.
(i) explain motion due to a uniform velocity in one direction
and a uniform acceleration in a perpendicular direction.
Equations of Motion
v = u + at • The symbols are:
– a acceleration (m/s2)
– s displacement (m)
s = (u + v)t – t time (s)
2 – u initial velocity (m/s)
– v final velocity (m/s)
s = ut + ½ at2 These equations work for speed
v2 = u2 + 2 as and distance as well
Deriving Equations of Motion
1. Write an equation for acceleration using the initial
velocity (u), the final velocity (v) and time (t).
2. Re-arrange the equation to give v. (This is the first
equation.)
3. Write an equation for displacement using
displacement = average velocity time. (This is the
second equation.)
4. Eliminate v from the second equation by substituting
it with (u + at) in equation 2. (This is the third
equation.)
5. Eliminate t from equation 3 by substituting t=(v-u)/a
(from equation 1) to get equation 4.
Example Question.
A sprinter (starting from rest) covers 4.0 m in the first
0.9s of a race. Find the acceleration, assuming it is
constant.
1. Write down what information the question gives and
requires:
u = 0.0 m/s (rest)
s = 4.0 m
t = 0.9 m/s
a=?
2. Look at the „suvat‟ equations and choose the one that
fits the quantities you have listed
u = 0.0 m/s (rest)
s = 4.0 m
v = u + at
t = 0.9 m/s
a=?
s = (u + v)t
3. Write down the
equation and substitute 2
the numbers into it. Don‟t
re-arrange as this would s = ut + ½ at2
make it more difficult to
check later. v2 = u2 + 2 as
s = ut + ½ at2
4. Finally work out the
4.0 = 0.0 + ½ a 0.92 answer, add the units,
check your working and
a = 2 4.0 /0.92 underline it when you are
sure it is right.
a = 9.88m/s2
The algorithm:
1. Write down what information the question
gives and requires.
2. Choose the equation that fits.
3. Write down the equation and substitute the
numbers into it.
4. Work out the answer, add the units, check
your working and underline it.
Example 1.
A car is travelling at 30 m/s and takes 10 seconds to
acceleration to a new speed of 35 m/s.
What is its acceleration ?
u = 30m/s
t = 10s
v = 35m/s
a=?
Use v = u + at
35 = 30 + a x 10
a = (35 – 30)/10
a = 10 m/s2
Example 2.
A brick falls off the top of a wall under construction and drops
into a bed of sand 14.5 m below.
It makes a dent in the sand 185 mm deep. What is:
a) The speed of the brick just before it hits the sand.
b) Its deceleration in the sand.
c) What would happen to a person undergoing that
deceleration?
a) The speed of the brick just before it hits the sand.
u = 0m/s, s=14.5m, a = 9.81m/s2, v = ?
Use v2 = u2 + 2as
= 0 + 2 x 9.81 x 14.5
= 284.5
v = 16.9 m/s
b) Deceleration in the sand:
u = 16.9m/s, v= 0m/s, s = 0.185m, a = ?
Use v2 = u2 + 2as.
0 = 284.5 + 2 x a x 0.185
Rearranging:
2 x a x 0.185 = - 284.5
a = -284.5
2 x 0.185
= - 769 m/s2
c) What would happen to a person undergoing that
deceleration?
769 m/s2 is about 77 g, quite sufficient to cause fatal
injury.
Linking Equations of Motion
to Velocity – Time Graphs
v • v = u + at
• Velocity at end = velocity
at start + (acceleration ×
u time)
m/s
velocity
t
Time (s)
Linking Equations of Motion
to Velocity – Time Graphs
• s = (u + v)t )
v Velocity
(m/s
2 u
u + v (m/s)
2
• Displacement = average
velocity × time
Time (s)
Linking Equations of Motion to
Velocity – Time Graphs
• We know that the area under
the graph is the displacement.
We can work this out adding v
the areas of the light rectangle
and the dark triangle.
v – u = at
• The area of the rectangle is u ×
u
t
• The area of the triangle = ½ ×
Speed (m/s)
acceleration × time × time.
• In code, area of the triangle = t
½ (at2).
• The total area, the
displacement covered is given Time (s)
by:
• s = ut + ½ at2